Number Property

Question

What is the remainder when 111×106×109 is divided by 15?

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Kinshook · Answer

Asked: What is the remainder when 111×106×109 is divided by 15?

111×106×109 mod 15 
= (105+6)*(105+1)(105+4) mod 15
= 6*1*4 mod 15
24 mod 15
(15+9) mod 15
9 mod 15

The remainder when 111×106×109 is divided by 15 = 9

100mitra · Answer

Very simple 
11 1  x 10 6  x 10 9  
Any multiple of 5, 15, 25 will either end with 5, or 0
With this collect last digit of above multiple
(1 x 6 x 9) = 54
15 x 3 = 45 
Thus 54 - 45 = 9
 Reminder is 9

IanStewart · Answer

It's actually a bit of a coincidence that you got the right answer here only looking at units digits (technically, if we're dividing by 15, you'd get the right answer 1/3 of the time this way, because you aren't considering the '3' in '15'). You can only use units digits to reliably find remainders if you're dividing specifically by 2, 5 or 10. You can see that units digits won't always give the correct remainder by 15 if you try a few random other examples -- e.g. if you try to find the remainder of 13*14 when dividing by 15, looking only at the units digits of 13 and 14 won't give you the right answer.

Units digits here do get us partway to an answer: the units digit of the product must be 4. I find it a bit easier now to think about dividing by 30 rather than by 15: if our units digit is 4, the remainder by 30 can only be 4, 14 or 24. But the product is also a multiple of 3, because it includes '111', which is a multiple of 3. So our remainder can only be 24 when we divide by 30, because 30q + 4 and 30q + 14 are not multiples of 3 (we're adding a multiple of 3 and a non-multiple of 3, which will never give us a multiple of 3). If we get a remainder of 24 when we divide by 30, so if our number is 30q + 24, then our number is also 30q + 15 + 9 = 15(something) + 9, and 9 is the remainder by 15.

Or we can reduce all the numbers in the product by any multiple of 15, since the 15s in our numbers will only affect the quotient when we divide by 15, not the remainder. Then we can just do the calculation with smaller numbers, which is exactly what kinshook did above. So we can replace 106 with 1, since 105 is a multiple of 15, and we can replace 111 with 6, and 109 with 4. We then just need the remainder when we divide 1*6*4 by 15, or 24 by 15, so the answer is 9.

100mitra · Answer

Thank you IanStewart This is an excellent learning and eye opener peice of knowledge.

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