Current Events > 15! = 13076743abc00

(+) Yeah! (+)

How do you find the digit values of a, b, and c without using a calculator?

multiply it out by hand

FLOUR posted...

How do you find the digit values of a, b, and c without using a calculator?

Use Google instead

does it even matter?

Just do a lucky guess. The odds you get it right are much, much higher than winning the lottery.

Since 15! has three factors of 5, and more than 3 factors of 2, it's certainly a multiple of 1000. So c = 0.

15! is also a multiple of 9, and 1 + 3 + 0 + 7 + 6 + 7 + 4 + 3 + a + b + 0 + 0 + 0= 31 + a + b which must be a multiple of 9. So 31 + a + b = 36 or 45 (a + b can't be higher than 14 because a and b are integers from 0 to 9 and if a + b = 23 then we need a or b to be over 9.) So a + b = 5 or 14.

15! is a multiple of 11, so 1 - 3 + 0 -7 + 6 - 7 + 4 - 3 + a - b +0 - 0 +0 = -9 + a - b is a multiple of 11. Again, since a and b are integers from 0 to 9, the only conceivable values of - 9 + a - b are -11 and 0.

Suppose - 9 + a - b = 0. Then a = b + 9. That can only be true if b = 0. But we don't have a fourth multiple of 5, so b can't be 0.

Suppose - 9 + a - b = -11. Then a - b = -2 so a = b -2 so a+ b = 2b - 2. 2b - 2 must be even, so it can't be 5. So a + b = 14. Now 2b - 2 = 14 which makes b = 8 and a = 6.

It probably would have been faster to just manually multiply.

Just with basic algebra, this level of reasoning is possible for Eviora. What do you think?

Eviora posted...

Since 15! has three factors of 5, and more than 3 factors of 2, it's certainly a multiple of 1000. So c = 0.

15! is also a multiple of 9, and 1 + 3 + 0 + 7 + 6 + 7 + 4 + 3 + a + b + 0 + 0 + 0= 31 + a + b which must be a multiple of 9. So 31 + a + b = 36 or 45 (a + b can't be higher than 14 because a and b are integers from 0 to 9 and if a + b = 23 then we need a or b to be over 9.) So a + b = 5 or 14.

15! is a multiple of 11, so 1 - 3 + 0 -7 + 6 - 7 + 4 - 3 + a - b +0 - 0 +0 = -9 + a - b is a multiple of 11. Again, since a and b are integers from 0 to 9, the only conceivable values of - 9 + a - b are -11 and 0.

Suppose - 9 + a - b = 0. Then a = b + 9. That can only be true if b = 0. But we don't have a fourth multiple of 5, so b can't be 0.

Suppose - 9 + a - b = -11. Then a - b = -2 so a = b -2 so a+ b = 2b - 2. 2b - 2 must be even, so it can't be 5. So a + b = 14. Now 2b - 2 = 14 which makes b = 8 and a = 6.

It probably would have been faster to just manually multiply.

Just with basic algebra, this level of reasoning is possible for Eviora. What do you think?

Well, I said "brown"...

Start by finding the factors of 15!
2^11 * 3^6 * 5^3 * 7^2 * 11 * 13

I can guarantee c=0 because 2^3 * 5^3 = 1000

Removing those factors: 2^8 * 3^6 * 7^2 * 11 * 13 = 13076743ab

a + b = 5 or 14 (1+3+7+6+7+4+3=31) as the entire number must be divisible by 9 (3^2), and the digits of any number divisible by 9 will add up to a multiple of 9

ab is also divisible by 4 as 2^2 is one of the factors, so we need a 2 digit number divisible by 4 whose digits add up to 5 or 14. The only 2 number who fit that bill are a=6 and b=8 as 68 is divisible by 4 but not 86.

Eviora posted...

Just with basic algebra, this level of reasoning is possible for Eviora. What do you think?

Overcomplicated solution. I give it 80/100.

Fair. However, your solution is incomplete since 32 is a 2 digit number divisible by 4 whose digits add up to 5. You did not go to the trouble of ruling it out.

Eviora posted...

Fair. However, your solution is incomplete since 32 is a 2 digit number divisible by 4 whose digits add up to 5. You did not go to the trouble of ruling it out.

Okay, 332 is not divisible by 8 but 368 is. Happy now?

That's a definite improvement. A proper proof would have gone through the list one by one to find then eliminate all the other possible two digit numbers with those properties, though. =p