Which numbers will make this number divisible by 99?

Question

What you need to do is find two digits. You place them at the beginning and end at the number i will state so that the number is divisible by 99. Let's call the 1st digit "a" and the 2nd digit "b". You basically need to do this: Find the values for a and b so that the number a8548736473674837483376b is divisible by 99. I know it looks hard, but how many options could there be with only a and b? Write you answer as the two digit number "ab".

Answer 1

Interesting problem. First, if the number is divisible by 99, then it must be divisible by both 11 and 9.

If the number is divisible by 9, then the sum of its digits must be divisible by 9. The sum of the digits is 121 + a + b. Therefore a + b = 5 (giving 126) or 12 (giving 135).

If the number is divisible by 11, then the sum of the odd digits minus the sum of the even digits must be divisible by 11. The sum of the odd digits is (a + 62). the sum of the even digits is (b + 59). So, (a + 62) - (b + 59) must be divisible by 11. Since a and b both have to be single digits, this leaves two possibilities: a = b - 3 or a = b + 8.

Combining the above two leaves only handful of combinations to test, and we see a = 1 and b = 4 as the only possible solution.

Answer 2

Any number divisible by 99 has to be divisible by 11 and 9.
For any number to be divisible by 9, the sum of the digits has to be a multiple of 9.The sum of the digits is 121 + a + b. Therefore a + b = 5 (giving 126) or 12 (giving 135).

If the number is divisible by 11, then the sum of the odd digits minus the sum of the even digits must be divisible by 11. The sum of the odd digits is (a + 62). the sum of the even digits is (b = 4 a+ 59). So, (a + 62) - (b + 59) must be divisible by 11. Since a and b both have to be single digits, this leaves two possibilities: a = b - 3 or a = b + 8.
(a+62)-(b+59) is divisible by 11
That means a+62-b-59 is divisible by 11
It implies that a-b+3 is divisible by 11.

Try a=b-3
If a=1, b=4.
Let's test it with the above equation

a-b+3 is divisible by 11
1-4+3 = 0 which is divisible by 11.
Therefore, your answer is a=1, b=4.
ab=14

Answer 3

To be divisible by 9, sum of digits has to be divisible by 9 also.
a+8+5+4+8+7+3+6+4+7+3+6+7+4+8+3+7+4+8+3+3+7+6+b
= a+b+121 ≡ a+b+4 (mod 9)

a+b≡5 (mod 9)

To be divisible by 11, (sum of even digits - sum of odd digits) has to be divisible by 11.

(a+5+8+3+4+3+7+8+7+8+3+6) - (8+4+7+6+7+6+4+3+4+3+7+b)
= a + 62 - 59 - b = a-b+3

a-b≡8 (mod 11)

Then trial and error:
I think there is only one solution: a=1,b=4
ab = 14

Answer 4

Hmmm... I'm going to have to think about this.

We know that a + the sum of digits of n + b must be a multiple of 9. And we know that the number must be divisible by 11. If we can satisfy both those things we are done...

Good job to zanti... I concur with his answer! Note (it would be different for a different number, but the method would work)

Answer 5

I have tried all 100 combinations for ab (00-99) in a TI-86 graphing calculator and none of the combinations divide evenly by 99.(WRONG)

Even the aforementioned proposed answer of ab = 14 does not work. (WRONG)

185487364736748374833764 / 99 = 187360874481563977.41414141414141(WRONG)

Note the repeating decimal of .41
Therefore i believe that there is no solution.(WRONG)

NOW WHEN I GO BACK AND TRY IT AGAIN THE SOLUTION OF ab = 14 WORKS. I GUESS THAT I NEED TO EAT BREAKFAST MORE OFTEN.

Answer 6

zanti3 is CORRECT, and anonymous with the TI-86 calculator is WRONG.

You can divide 1854 87364 73674 83748 33764 by 9 and then by 11 with pencil and paper, to get 18 73609 74481 56401 49836 remainder 0.

Answer 7

a854 8736 4736 7483 7483 376b
8 6 2 4 4 7===4
854 8736 4736 7483 7483 376mod9 = 4
854 8736 4736 7483 7483 376mod11 = 3
The required pair is
ab mod9 = 5,
4854 8736 4736 7483 7483 3761mod 11 = 3
1854 8736 4736 7483 7483 3764mod99 = 0

n = 14

Answer 8

There could be 100 different options