Calculation Tips & Tricks

Calculation Tips & Tricks
Divisibility Rules
Why do these 'rules' work? - Dr. Rob
Divisibilidad por 13 y por números primos (13,17,19...)
- en español, de la lista SNARK
From the Archives of the Math Forum's Internet project Ask Dr. Math - our thanks to Ethan 'Dr. Math' Magness, Steven 'Dr. Math' Sinnott, and, for the explanation of why these rules work, Robert L. Ward (Dr. Rob).
Dividing by 3
Add up the digits: if the sum is divisible by three, then the number is as well. Examples:
1. 111111: the digits add to 6 so the whole number is divisible by three.
2. 87687687. The digits add up to 57, and 5 plus seven is 12, so the original number is divisible by three.
Why does the 'divisibility by 3' rule work?
From: Dr. Math
To: Kevin Gallagher
Subject: Re: Divisibility of a number by 3
As Kevin Gallagher wrote to Dr. Math
On 5/11/96 at 21:35:40 (Eastern Time),
>I'm looking for a SIMPLE way to explain to several very bright 2nd
>graders why the divisibility by 3 rule works, i.e. add up all the
>digits; if the sum is evenly divisible by 3, then the number is as well.
>Thanks!
>Kevin Gallagher
The only way that I can think of to explain this would be as follows: Look


at a 2 digit number: 10a+b=9a+(a+b). We know that 9a is divisible by 3, so 10a+b will be divisible by 3 if and only if a+b is. Similarly, 100a+10b+c=99a+9b+(a+b+c), and 99a+9b is divisible by 3, so the total will be iff a+b+c is.
This explanation also works to prove the divisibility by 9 test. It clearly originates from modular arithmetic ideas, and I'm not sure if it's simple enough, but it's the only explanation I can think of.
Doctor Darren, The Math Forum
Check out our web site - http://mathforum.org/dr.math/
Dividing by 4
Look at the last two digits. If the number formed by its last two digits is divisible by 4, the original number is as well.
Examples:
3. 100 is divisible by 4.
4. 1732782989264864826421834612 is divisible by four also, because 12 is divisible by four.
Dividing by 5
If the last digit is a five or a zero, then the number is divisible by 5.
Dividing by 6
Check 3 and 2. If the number is divisible by both 3 and 2, it is divisible by 6 as well.
Robert Rusher writes in:
Another easy way to tell if a [multi-digit] number is divisible by six . . .
is to look at its [ones digit]: if it is even, and the sum of the [digits] is
a multiple of 3, then the number is divisible by 6.
Dividing by 7
To find out if a number is divisible by seven, take the last digit, double it, and subtract it from the rest of the number.
Example: If you had 203, you would double the last digit to get six, and subtract that from 20 to get 14. If you get an answer divisible by 7 (including zero) , then the original number is divisible by seven. If you don't know the new number's divisibility, you can apply the rule again.
Matthew Correnti describes this method:
If you do not know if a two-digit number, call it ab, is
divisible
by 7, calculate 2a + 3b. This will yield a smaller number, and if
you do the process enough times you will eventually -- if the
number ab is divisible by 7 -- end up with 7.

You can use a similar method if you have a three-digit number abc:
take the digit a and multiply it by 2, then add it to the number bc,
giving you 2a + bc; repeat and reduce until you recognize the
result's divisibility by seven. With a four-digit number abcd, take
the digit a and multiply by 6, then add 6a to bcd giving. This
usually gives you a three-digit number; call it xyz. Take that x
and, as described previously, multiply x by two and add to yz
(i.e., 2x + yz). Again, repeat and reduce until you recognize the
result's divisibility by seven.
Ahmed Al Harthy writes in:
To know if a number is a multiple of seven or not, we can use also
3 coefficients (1 , 2 , 3). We multiply the first number starting
from the ones place by 1, then the second from the right by 3,
the third by 2, the fourth by -1, the fifth by -3, the sixth by -2,
and the seventh by 1, and so forth.

Example: 348967129356876.

6 + 21 + 16 - 6 - 15 - 6 + 9 + 6 + 2 - 7 - 18 - 18 + 8 + 12 + 6 = 16
means the number is not multiple of seven.

If the number was 348967129356874, then the number is a multiple of seven
because instead of 16, we would find 14 as a result, which is a multiple of 7.

So the pattern is as follows: for a number onmlkjihgfedcba, calculate

a + 3b + 2c - d - 3e - 2f + g + 3h + 2i - j - 3k - 2l + m + 3n + 2o.

Example: 348967129356874.

Below each digit let me write its respective figure.

3 4 8 9 6 7 1 2 9 3 5 6 8 7 6
2 3 1 -2 -3 -1 2 3 1 -2 -3 -1 2 3 1

(3×2) + (4×3) + (8×1) + (9×-2) + (6×-3) + (7×-1) +
(1×2) + (2×3) + (9×1) + (3×-2) + (5×-3) + (6×-1) +
(8×2) + (7×3) + (6×1) = 16 -- not a multiple of seven.
Another visitor observes:
Here is one formula for seven...

3X + L

L = last digit
X = everything in front of last digit.

All numbers that are divisible by seven have this in common.
There are no exceptions.

For example, 42: 3(4) + 2 = 14.
Seven divides into 14, so it divides into 42.

Next example, 105: 3(10) + 5 = 35.
Seven divides into 35, so it divides into 105.

Here is another formula for seven:

4X - L

When using this formula, if you get zero, seven or a multiple of seven,
the number will be divisible by seven.

For example, 56: 4(5) - 6 = 14.
Seven divides into 14, so it divides into 56.

Next example, 168: 16(4) - 8 = 56.
Seven divides into 56, so it divides into 168.

Similarly:
The formula for 2 is 2X + L
The formula for 3 is 4X + L
The formula for 4 is 6X + L
The formula for 5 is 5X + L
The formula for 6 is 2X + L and 4X + L -- in other words, the formulas for 2 and 3
must work before the number is divisible by 6.
The formula for 9 is X + L
The formula for 11 is X - L
The formula for 12 is 2X - L
The formula for 13 is 3X - L
The formula for 14 is 4X - L and 2X + L -- in other words, the formulas for 7 and 2
must work before the number is divisible by 14.
The formula for 17 is 7X - L
The formula for 21 is X - 2L
The formula for 23 is 3X - 2L
The formula for 31 is X - 3L
Sara Heikali explains this way to test a number with three or more digits for divisibility by seven:
1. Write down just the digits in the tens and ones places.
2. Take the other numbers to the left of those last two digits,
and multiply them by two.
3. Add the answer from step two to the number from step one.
4. If the sum from step three is divisible by seven, then the
original number is divisible by seven, as well. If the sum is
not divisible by seven, then the original number is not
divisible by seven.

For example, if the number we are testing is 112, then
1. Write down just the digits in the tens and ones places: 12.
2. Take the other numbers to the left of those last two digits,
and multiply them by two: 1 × 2 = 2.
3. Add the answer from step two to the number from step one:
12 + 2 = 14.
4. Fourteen is divisible be seven. Therefore, our original
number, one hundred twelve, is also divisible by seven.
See also Explaining the Divisibility Rule for 7 in the Dr. Math archives.
Dividing by 8
Check the last three digits. Since 1000 is divisible by 8, if the last three digits of a number are divisible by 8, then so is the whole number.
Example: 33333888 is divisible by 8; 33333886 isn't.
How can you tell whether the last three digits are divisible by 8? Phillip McReynolds answers:
If the first digit is even, the number is divisible by 8 if the last two digits are. If the first digit is odd, subtract 4 from the last two digits; the number will be divisible by 8 if the resulting last two digits are. So, to continue the last example, 33333888 is divisible by 8 because the digit in the hundreds place is an even number, and the last two digits are 88, which is divisible by 8. 33333886 is not divisible by 8 because the digit in the hundreds place is an even number, but the last two digits are 86, which is not divisible by 8.
Sara Heikali explains this test of divisibility by eight for numbers with
three or more digits:
1. Write down the units digit of the original number.
2. Take the other numbers to the left of the last digit,
and multiply them by two.
3. Add the answer from step two to the number from step one.
4. If the sum from step three is divisible by eight, then the
original number is divisible by eight, as well. If the sum is
not divisible by eight, then the original number is not
divisible by eight.

For example, if the number we are testing is 104, then
1. Write down just the digits in ones place: 4.
2. Take the other numbers to the left of that last digit,
and multiply them by two: 10 × 2 = 20.
3. Add the answer from step two to the number from step one:
4 + 20 = 24.
4. Twenty-four is divisible be eight. Therefore, our original
number, one hundred and four, is also divisible by eight.
Dividing by 9
Add the digits. If that sum is divisible by nine, then the original number is as well.
Dividing by 10
If the number ends in 0, it is divisible by 10.
Dividing by 11
Let's look at 352, which is divisible by 11; the answer is 32. 3+2 is 5; another way to say this is that 35 -2 is 33.
Now look at 3531, which is also divisible by 11. It is not a coincidence that 353-1 is 352 and 11 × 321 is 3531.
Here is a generalization of this system. Let's look at the number 94186565.
First we want to find whether it is divisible by 11, but on the way we are going to save the numbers that we use: in every step we will subtract the last digit from the other digits, then save the subtracted amount in order. Start with
9418656 - 5 = 9418651 SAVE 5
Then 941865 - 1 = 941864 SAVE 1
Then 94186 - 4 = 94182 SAVE 4
Then 9418 - 2 = 9416 SAVE 2
Then 941 - 6 = 935 SAVE 6
Then 93 - 5 = 88 SAVE 5
Then 8 - 8 = 0 SAVE 8
Now write the numbers we saved in reverse order, and we have 8562415, which multiplied by 11 is 94186565.
________________________________________
Here's an even easier method, contributed by Chis Foren:
Take any number, such as 365167484.
Add the first, third, fifth, seventh,.., digits.....3 + 5 + 6 + 4 + 4 = 22
Add the second, fourth, sixth, eighth,.., digits.....6 + 1 + 7 + 8 = 22
If the difference, including 0, is divisible by 11, then so is the number.
22 - 22 = 0 so 365167484 is evenly divisible by 11.
See also Divisibility by 11 in the Dr. Math archives.
Dividing by 12
Check for divisibility by 3 and 4.
Dividing by 13
Here's a straightforward method supplied by Scott Fellows:
Delete the last digit from the given number. Then subtract nine times the deleted digit from the remaining number. If what is left is divisible by 13, then so is the original number.
________________________________________
Rafael Ando contributes:
Instead of deleting the last digit and subtracting it ninefold from the remaining number (which works), you could also add the deleted digit fourfold. Both methods work because 91 and 39 are each multiples of 13.
For any prime p (except 2 and 5), a rule of divisibility could be "created" using this method:
2. Find m, such that m is the (preferably) smallest multiple of p that ends in either 1 or 9.
3. Delete the last digit and add (if multiple ends in 9) or subtract (if it ends in 1) the deleted digit times the integer nearest to m/10. For example, if m = 91, the integer closest to 91/10 = 9.1 is 9; and for 3.9, it's 4.
4. Verify if the result is a multiple of p. Use this process until it's obvious.
Example 1: Let's see if 14281581 is a multiple of 17.
In this case, m = 51 (which is 17×3), so we'll be deleting the last number and subtracting it fivefold.
1428158 - 5×1 = 1428153
142815 - 5×3 = 142800
14280 - 5×0 = 14280
1428 - 5×0 = 1428
142 - 5×8 = 102
10 - 5×2 = 0, which is a multiple of 17, so 14281581 is multiple of 17.
Example 2: Let's see if 7183186 is a multiple of 46.
First, note that 46 is not a prime number, and its factorization is 2×23. So, 7183186 needs to be divisible by both 2 and 23. Since it's an even number, it's obviously divisible by 2.
So let's verify that it is a multiple of 23:
m = 3×23 = 69, which means we'll be adding the deleted digit sevenfold.
718318 + 7×6 = 718360
71836 + 7×0 = 71836
7183 + 7×6 = 7225
722 + 7×5 = 757
75 + 7×7 = 124
12 + 7×4 = 40
4 + 7×0 = 4 (not divisible by 23), so 7183186 is not divisible by 46.
Note that you could've stopped calculating whenever you find the result to be obvious (i.e., you don't need to do it until the end). For example, in example 1 if you recognize 102 as divisible by 17, you don't need to continue (likewise, if you recognized 40 as not divisible by 23).
The idea behind this method it that you're either subtracting m×(last digit) and then dividing by 10, or adding m×(last digit) and then dividing by 10.
________________________________________
Jeremy Lane adds:
It may be noted that while applying these rules, it is possible to loop among numbers as results.
Example: Is 1313 divisible by 13?
Using the procedure given we take 13×3 and obtain 39. This multiple ends in 9 so we add four-fold the last digit.
131 + 4×3 = 143
14 + 4×3 = 26
2 + 4×6 = 26
...
Example: Is 1326 divisible by 13?
Using the procedure given we take 13×7 = 91. This is not the smallest multiple, but it does show looping. The smaller multiple does loop at 39 as well. There are some examples where we would still need to recognize certain multiples. So we subtract nine-fold the last digit.
132 - 9×6 = 78
7 - 9×8 = -65 (factor out -1)
6 - 9×5 = -39 (again factor out -1)
3 - 9×9 = -78 (factor out -1)
This only occurs though if the number does happen to be divisible by the prime divisor. Otherwise, eventually you will have a number that is less than the prime divisor.
________________________________________
And here's a more complex method that can be extended to other formulas:
1 = 1 (mod 13)
10 = -3 (mod 13) (i.e., 10 - -3 is divisible by 13)
100 = -4 (mod 13) (i.e., 100 - -4 is divisible by 13)
1000 = -1 (mod 13) (i.e., 1000 - -1 is divisible by 13)
10000 = 3 (mod 13)
100000 = 4 (mod 13)
1000000 = 1 (mod 13)
Call the ones digit a, the tens digit b, the hundreds digit c, ..... and you get:
a - 3×b - 4×c - d + 3×e + 4×f + g - .....
If this number is divisible by 13, then so is the original number.
You can keep using this technique to get other formulas for divisibility for prime numbers. For composite numbers just check for divisibility by divisors.
Dividing by 14
Sara Heikali builds on her divisibility test for seven:
How can you know if a number with three or more digits
is divisible by the number fourteen?

Check if the last digit of the original number is odd or
even. If the number is odd, then the number is not
divisible by fourteen. If the number is even, then apply
the divisibility rule for seven.

(Keep in mind, the odd and even test is to see if the number
is divisible by two.) If the original even number is
divisible by seven, then it is also divisible by fourteen.
If the original even number is not divisible by seven, it
is not divisible by fourteen.





Multiplication Tips
Multiplying by five
• Jenny Logwood writes: Here is an easy way to find an answer to a 5 times question.
If you are multiplying 5 times an even number: halve the number you are multiplying by and place a zero after the number. Example: 5 × 6, half of 6 is 3, add a zero for an answer of 30. Another example: 5 × 8, half of 8 is 4, add a zero for an answer of 40.
If you are multiplying 5 times an odd number: subtract one from the number you are multiplying, then halve that number and place a 5 after the resulting number. Example: 5 × 7: -1 from 7 is 6, half of 6 is 3, place a 5 at the end of the resulting number to produce the number 35. Another example: 5 × 3: -1 from 3 is 2, half of 2 is 1, place a 5 at the end of this number to produce 15.
• Doug Elliott adds: To square a number that ends in 5, multiply the tens digit by (itself+1), then append 25. For example: to calculate 25 × 25, first do 2 × 3 = 6, then append 25 to this result; the answer is 625. Other examples: 55 x 55; 5 × 6 = 30, answer is 3025. You can also square three digit numbers this way, by starting with the the first two digits: 995 x 995; 99 × 100 = 9900, answer is 990025.
Multiplying by nine
• Diana Grinwis says: To multiply by nine on your fingers, hold up ten fingers - if the problem is 9 × 8 you just put down your 8 finger and there's your answer: 72. (If the problem is 9 × 7 just put down your 7 finger: 63.)
• Laurie Stryker explains it this way: When you are multiplying by 9, on your fingers (starting with your thumb) count the number you are multiplying by and hold down that finger. The number of fingers before the finger held down is the first digit of the answer and the number of finger after the finger held down is the second digit of the answer.
Example: 2 × 9. your index finder is held down, your thumb is before, representing 1, and there are eight fingers after your index finger, representing 18.
• Polly Norris suggests: When you multiply a number times 9, count back one from that number to get the beginning of your product. (5 × 9: one less than 5 is 4).
To get the rest of your answer, just think of the add fact families for 9:
1 + 8 = 9 2 + 7 = 9 3 + 6 = 9 4 + 5 = 9
8 + 1 = 9 7 + 2 = 9 6 + 3 = 9 5 + 4 = 9
5 × 9 = 4_. Just think to yourself: 4 + _ = 9 because the digits in your product always add up to 9 when one of the factors is 9. Therefore, 4 + 5 = 9 and your answer is 45! I use this method to teach the "nines" in multiplication to my third graders and they learn them in one lesson!
Tamzo explains this a little differently:
1. Take the number you are multiplying 9 by and subtract one. That number is the first number in the solution.
2. Then subtract that number from nine. That number is the second number of the solution.
Examples:
4 * 9 = 36
1. 4-1=3
2. 9-3=6
3. solution = 36
8 * 9 = 72
4. 8-1=7
5. 9-7=2
6. solution = 72
5 * 9 = 45
7. 5-1=4
8. 9-4=5
9. solution = 45
• Sergey writes in: Take the one-digit number you are multipling by nine, and insert a zero to its right. Then subtract the original number from it.
For example: if the problem is 9 * 6, insert a zero to the right of the six, then subtract six:
9 * 6 = 60 - 6 = 54
Multiplying a 2-digit number by 11
• A tip sent in by Bill Eldridge: Simply add the first and second digits and place the result between them.
Here's an example using 24 as the 2-digit number to be multiplied by 11: 2 + 4 = 6 so 24 × 11 = 264.
This can be done using any 2-digit number. (If the sum is 10 or greater, don't forget to carry the one.)
Multiplying any number by 11
• Lonnie Dennis II writes in:
Let's say, for example, you wanted to multiply 54321 by 11. First, let's look at the problem the long way...
54321
x 11
________________________________________54321
+ 543210
________________________________________= 597531
Now let's look at the easy way...
11 × 54321
= 5 4+5 4+3 3+2 2+1 1
= 597531
Do you see the pattern? In a way, you're simply adding the digit to whatever comes before it.
But you must work from right to left. The reason I work from right to left is that if the numbers, when added together, sum to more than 9, then you have something to carry over.
Let's look at another example...
11 × 9527136
Well, we know that 6 will be the last number in the answer. So the answer now is
???????6.
Calculate the tens place: 6+3=9, so now we know that the product has the form
??????96.
3+1=4, so now we know that the product has the form
?????496.
1+7=8, so
????8496.
7+2=9, so
???98496.
2+5=7, so
??798496.
5+9=14.
Here's where carrying digits comes in: we fill in the hundred thousands place with the ones digit of the sum 5+9, and our product has the form
?4798496.
We will carry the extra 10 over to the next (and final) place.
9+0=9, but we need to add the one carried from the previous sum: 9+0+1=10.
So the product is 104798496.
Multiplying by thirteen
• Fourth grader Mariam Labib has a trick for multiplication by 13.
Put the tens digit on the left, the unit number on the right, add them up together in the middle. Then add double the number to the previous result.
For example: 13 × 22
Step 1: (2 × 100) + 2 + [(2 + 2) × 10] = 242.
Step 2: 22 × 2 = 44.
Answer: 242 + 44 = 286.
If the two digits sum to more than ten, then you carry the one to add it to the number on the left and continue.
For example: 13 × 65
Step 1: (6 × 100) + 5 + [(6 + 5) × 10] = 715.
Step 2: 65 × 2 = 130.
Answer: 715 + 130 = 845.
Multiplying by sixteen
• Ibrahim Labib offers a quick way to find an answer when multiplying by 16.
First, multiply the number in question by 10. Then multiply half the number by 10. Then add those two results together with the number itself to get your final answer.
For example: 16 × 24
Step 1: 24 × 10 = 240
Step 2: (24 × 1/2) × 10 = 12 × 10 = 120
Step 3: add steps 1 and 2 and the number = 240 + 120 + 24 = 384




Multiplication Tips
Multiplying 2-digit numbers (same 1st digit)
• Fifth grader Mark Labib writes: My trick is about multiplying any 2 digit number by any other 2 digit number as long as they have the same tens digit. For example,
• 23 × 26 OR 42 × 47 OR 51 × 59
• As it gets harder with the larger numbers, I will start with multiplying two numbers in the twenties. For example,
• 23 × 27
Step 1. Add the 3 (from the 23) to the 27: 3 + 27 = 30
Step 2. Multiply the result by 10: 30 × 10 = 300
Step 3. Double that result: 2 × 300 = 600
Step 4. Multiply 3 (from the 23) and 7 (from the 27): 3 × 7 = 21
Step 5. Add these last two products: 600 + 21 = 621
• For two numbers in the thirties, when you come to step 3, triple the result; for the forties, quadruple it (just by doubling the number twice); for the fifties, multiply by 5 (using the tips for multiplication by 5: divide the number by 2 and multiplying it by ten); and so on.




Spring Mathematics Olympiad Problems
Organized by
the Palace of Youth Creativity
St. Petersburg, Russia

Back to H.S. Student Center || K-12 Math Problems

These problems are intended for students studying upper level secondary mathematics. Nearly all of them are new and original, proposed by Russian teachers especially for the Olympiad.
1. One bacterium divides into two new ones. Then every second, one of the bacteria also splits in two, and soon there are 1995 bacteria.
Prove that one of them lives for at least 995 seconds without any division.
________________________________________
2. A sequence of natural numbers (Xn ) is formed in the following way:
3. X = 19 X = 95 X = LCM ( X , X ) + X
4. 1 2 n+2 n+1 n n
5.
Find GCD ( X1995 , X1996 ).
[LCM - least common multiple
GCD - greatest common divisor]
________________________________________
6. In a country there live 'knights' and 'knaves.' The 'knaves' always lie and the 'knights' always tell the truth - unless they make a mistake.
1995 knights and knaves are sitting around a table. Each says that he is sitting between a knight and a knave, but two knights are mistaken. How many knaves are seated at the table?
________________________________________
7. Divide a square into 7 polygons such that any two like polygons will have no common edge, and any two that are not the same shape will have a common edge. (For any polygon there must exist another with the same shape.)
________________________________________
8. The Mad Tea-Party:
The Hatter, the March Hare, and the Dormouse sit at a round table. On the table are 12 cups full of tea.
Every day at six o'clock, each of the table-companions moves two places to his right or left (if the place is free) and drinks all the tea at that place if there is any in his cup. After the place-changing, Alice fills one of the cups. Prove that Alice can keep at least 6 cups full.
________________________________________
9. Prove that there are infinitely many integers that are not equal to the sum of two squares of integers, but that are equal to the sum of three cubes of integers.
________________________________________
10. Dazzy fishes for more than a week. He experiences three different levels of success. On a good day he catches 9 fish. On a bad day he catches only 5 fish. On other days he catches 7 fish.
In all, Dazzy catches 53 fish. How many bad days does he have?
11. Is it possible to put the numbers from 1 to 64 in the squares of a chess board in such a way that the sum of the numbers in any three squares arranged in the following pattern will be odd?
12. ---
13. | |
14. -------
15. | | |
-------
________________________________________
16. Tyrannosaurus Rex and Tyrannosaurus Glipp are playing a game. Each player in turn puts an X or an O in any free square of an 11x11 checkerboard.
The winner of the game is the player who first puts three equal marks in a line, horizontally, vertically, or diagonally:
XXX or O X
O or X
O X
Tyrannosaurus Rex starts. In an errorless game, who wins?
________________________________________
17. Ann, Ben, Can, and Den think of one natural number.
Ann says it consists of two digits.
Ben says it is a divisor of 150.
Can says it is not 150.
Den says it is divisible by 25.
Which one of them is not telling the truth?
________________________________________
18. All the digits from 1 to 9 are written on a blackboard. Pete erases some of the digits and writes the product of the numbers he has erased. In addition, he can write new digits. After some of these operations only one digit remains on the blackboard.
Prove that the remaining digit is 0.
________________________________________
19. Winnie-the-Pooh and Piglet are celebrating their common birthday. Every guest gives them a can of honey and a can of condensed milk. Pooh leaves some of the cans for Piglet, and others he keeps for himself.
When Pooh has eaten all of his honey, the number of full cans he has left is equal to the total number of Piglet's cans. Piglet has 10 cans of milk. How many cans of honey did Pooh eat?
________________________________________
20. In this game, there are 100 coins on one scale, and 200 coins on another scale. Each of two players in turn takes one or more of the coins from one of the scales. The player who makes the number of coins on the two scales equal loses.
Scrooge and Glomgold play a game. Scrooge makes the first move. In an errorless game, who wins?
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21. The calculator 'ComeOn-96' has the buttons '+1', '-1', '+4', and '-4'. Whenever the result of an operation is divisable by 4, the calculator explodes! ;-)
Begin with 1 on the calculator, and prove that it is impossible to arrive at an answer of 2 when using exactly 1996 operations.
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22. In a tribe of cannibals there are 20 persons. Each person likes some of his or her fellow-tribesmen and does not like others. Each day all of the people who are liked by more than half the tribe become dinner.
Prove that on the 10th day after the first cannibal has been eaten, because no person is liked by more than half of the other cannibals there is no dinner.
________________________________________
23. Prove that it is possible to write all 10-digit numbers that consist of only ones and twos in a circle in such a way that every two neighbouring numbers will differ by only one digit (e.g. 22211 and 22111, or 12121 and 12221).
________________________________________
24. Let S be a square, Q be the perimeter of the square, and P be the perimeter of a quadrilateral T inscribed within S such that each of its vertices lies on a different edge of S.
What is the smallest possible ratio of P to Q?
________________________________________
25. What is the greatest number of kings that can be placed on a chess board such that every king will win only one square or no square? (A king wins all eight neighbouring squares, as shown below.)
26. - - -
27. |*|*|*|
28. - - -
29. |*|K|*|
30. - - -
31. |*|*|*|
32. - - -
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33. All the natural numbers from 2 to 3500 are written on the blackboard. Players in turn can erase both any non-prime number and all the numbers that are not co-prime to it. A player wins if, after his turn, only prime numbers remain on the board, so that another player cannot take a turn.
T-Rex and T-Glipp play a game. Who will win if T-Rex begins and neither makes a mistake?
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34. In the country of OneWay, some cities are connected by roads. Cars may drive in only one direction on roads, but pedestrians can go in both directions.
From Return-City one can go to any other city by car, but in only one direction. If it is possible to leave a city and return to the same city by car, the car will go through Return-City.
Two strangers leave GetOff-City from different sides. They are both pedestrians and neither can visit any city more than once. Some time later, they meet again. Prove that one of them must have visited Return-City.


Mathematics Olympiad Problems, 1996
Vologda, Russia - for Grades 5-11
by Victor Guba

Back to H.S. Student Center || K-12 Math Problems
These problems come from Professor Victor Guba, formerly of the Vologda Pedagogical University (Russia), and currently of Vanderbilt University. To request solutions or other information about the olympiad, please visit his web site
http://www.math.vanderbilt.edu/~msapir/
1. A zoo has several ostriches and several giraffes. They have 30 eyes and 44 legs. How many ostriches and how many giraffes are in the zoo?
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2. Is there a configuration of 6 points and several closed intervals in the plane such that any point is joined exactly with a) 3 of the points, 4 of the points? (The intervals do not have inner intersections.)
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3. Is it true that n^3+5n-1 is prime for any natural n?
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4. A string of 1996 digits begins with the number 6. Any number formed by two consecutive digits is divisible by 17 or 23. What is the last digit?
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5. There are 3 boxes with balls inside them. They are labeled 'two whites', "two blacks', and 'black & white' to show the contents of the boxes. Someone changes the labels to make them all false. Find the contents of each box by removing only one of the balls.
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6. Divide a cube into 8 equal parts using a) four, b) three planar cuts.
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7. Is it true that
a. for any sequence of 5 different numbers, one may choose an increasing or decreasing subsequence of three numbers?
b. for any sequence of 9 different numbers one may choose an increasing or decreasing subsequence of four numbers?
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8. Is it possible to fill in a 5x5 table using numbers such that the sum of all them is positive, but the sum of any 4 numbers forming a 2x2 sub-table is negative?
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9. Solve the following multiplication example
10. ***
11. **
12. -----
13. ****
14. ****
15. -----
16. *****
where all the asterisks are prime digits (2,3,5,7).
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17. Is there a closed polygonal line that intersects any of its edges exactly once and consists of a) six, b) seven edges?
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18. Is it true that for any of 100 integer numbers, there may be chosen a) 15, b) 16 of them such that the difference between any two of them is divisible by 7?
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19. There is a straight-line scratch on a chess board. What is the greatest possible number of scratched squares?
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20. Is it possible to fill 25 squares of squared paper such that each of the filled squares has a) an even and nonzero, b) an odd number of filled neighbors?
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21. Find 3 numbers such that each of them is a square of the difference of the two others.
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22. Two players compose a 2000-digit number by taking turns at writing a digit from 1 to 5. To win the game, Player 2 must cause the result to be divisible by 9, so Player 1 attempts to prevent this from occurring. If a) k = 10, b) k = 15, who will win?
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23. A convex quadrilateral is divided diagonally into 4 triangles. The areas of three of these triangles, proceeding clockwise, are 1, 2, and 3. What is the area of the fourth triangle?
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24. The game 'Sea Battle' takes place on an 8x8 board. What is the least number of shots needed to hit a ship 4x1 squares in size?
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25. Is there an infinite increasing geometric progression such that the first 100 elements of the sequence but none of the rest of the elements are integers?
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26. Is it possible to find a rectangle ABCD and a point M on a plane such that the distances from M to A,B,C,D respectively are
a. 1,4,8,7
b. 1,2,3,4
c. 1,2,4,3 ?
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27. Present 100 as a sum of positive integers such that their product is maximal.
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28. Find the largest possible number n within which, for any permutation of the numbers 1,2,...,100, there are 10 consecutive numbers (with respect to a permutation) such that their sum is greater than or equal to n.
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29. A chess tournament involves only experts and grand masters. The number of experts is three times the number of grand masters. Only one game is played between each pair of players, with 1 point being awarded for a win, 0 points for a loss, and 1/2 point for a draw. The sum of the points of all the experts is 1.2 times greater than the sum of points of all the grand masters. How many participants does the tournament have?
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30. How many solutions does the equation sin x = x/9 have?
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31. Find the number of 10-digit numbers that involve only the numbers 2 and 3 but do not have two 3's next to each other.
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32. Is the number [(44+\sqrt{1996})^{100}] odd or even?
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33. Quadrilateral ABCD is inscribed in a circle. Let Hi (i=1,2,3,4) be the point of intersection of the three altitudes of triangles BCD, ACD, ABD, and ABC, respectively. Prove that quadrilateral H1H2H3H4 is equal to quadrilateral ABCD.
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34. A circle is inscribed in a triangle ABC. The circle touches AB and BC at points E and F, respectively. Angle bisector A intersects line EF at point K. Show that angle CKA is a right angle.
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The sum of the lengths of a system of intervals of a line is less than 1. Given a set of 100 points on the line, show that they can be shifted by a distance less than 50 such that no images of these points under the shift belong to the system of intervals.