If m is a digit between 0 and 9, what is the remainder when the 7

Question

If m is a digit between 0 and 9, what is the remainder when the 7 digit integer 1,8m7,359 is divided by 16?

A. 1
B. 3
C. 7
D. 15
E. ﻿﻿Cannot be determined from the given information

Nidzo · Accepted Answer

In this instance the value of  m  is irrelevant. When looking at the divisibility rule of numbers which are exponents of 2, they follow the following pattern: a number is divisible by 2 if the last digit is divisible by 2, a number is divisible by 4 if the last two digits are divisible by 4, a number is divisible by 8 if the last three digits are divisible by 8 and a number is divisible by 16 if the last four digits are divisible by 8. Another fact is that whatever the remainder is when testing these rules, will be the remainder when the entire number is divided by that number.

ie.  \frac{555325}{4} . To test divisibility:  \frac{25}{4} = 6  with a remainder of  1 . This means  555325  is not divisible by  4 , and that  555325   divided by  4   leaves a remainder of  1 .

With this in mind,  1,8m7,359  divided by  16 : we take the last four digits and divide by  16

\frac{7359}{16} = 459  with a remainder of  15

ANSWER D

Prince1890Sharma · Answer

Hello, Can anyone please explain the answer to this question?

DmitryFarberMPrep · Answer

The first thing I tried to figure out here was how many of the digits really mattered. If I want to divide by 2, for instance, I only care about the units digit, since 10, 100, etc. are all multiples of 2, so they'll leave no remainder. Same idea for 4: if I want to divide by 4, I know it will go into everything from 100 on up, so I just care about the tens and units. If the last 2 digits make a multiple of 4, the whole thing is a multiple of 4.

Extending this pattern, I can notice that 16 = 2^4. This means that for a number to be a multiple of 16, it needs four 2's in its factorization. Each time I go up by a power of 10, I get another 2. So 10 is divisible by 2, 100 is divisible by 4, 1,000 is divisible by 8, and 10,000 is divisible by 16. In other words, the first three digits (1,8,m) don't matter, since that part will already be a multiple of 16. If you can recognize all this within the first 45 seconds, then you have plenty of time to deal with the rest.

From this point, I just have to divide 7359 by 16. Long division is probably the simplest way, but if you recognize that 7200 is a multiple of 16 (9*8*100), then all that's left is 159. That's one less than 160. Since it's one below a multiple of 16, it's 15 above the previous multiple.

MT1302 · Answer

Hey, there is a Divisibility rule for 2 which states that :

If a no. is divisible by 2 then  last digit  for that no. should be divisible by 2 . For eg : 1222 ---> Last digit is 2 ,
Hence, 1222 is divisible by 2.

If a no. is divisible by 4 then  last 2 digits  for that no. should be divisible by 4 . For eg : 1220 ---> Last 2 digits are 20 .
Since, 20 is Divisible by 4 Hence, 1220 is divisible by 4.

If a no. is divisible by 8 then  last 3 digits  for that no. should be divisible by 8 . For eg : 1800 ---> Last 3 digits are 800 .
800 is Divisible by 8. Hence, 1800 is divisible by 8.

If a no. is divisible by 16 then  last 4 digits  for that no. should be divisible by 16. For eg : 11600 ---> Last 4 digits are 1600 .
1600 is Divisible by 16. Hence, 11600 is divisible by 16.

So, in this question, value of  m  doesn't matters because we can directly get the remainder by dividing 7359 by 16.

Hence,  D  is the answer.

Hope it helps .

BrushMyQuant · Answer

What is the remainder when the 7 digit integer 1,8m7,359 is divided by 16

Remainder of a number by 16 = Remainder of last 4 digits of the number by 16
=> Remainder of 1,8m7,359 by 16= Remainder of 7359 by 16 = 15

So,  Answer will be D 
Hope it helps!

Watch the following video to MASTER Remainders

https://www.youtube.com/watch?v=P0S6j2uTaj4

bumpbot · Answer

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If m is a digit between 0 and 9, what is the remainder when the 7

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If m is a digit between 0 and 9, what is the remainder when the 7

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