What is the largest four digit number when it is divided by 4, 6 and 8

Question

What is the largest four digit number when it is divided by 4, 6 and 8 leaves remainders 3, 5 and 7 respectively.
a) 9872
b) 9997
c) 9936
d) 9983
e) 9883

broall · Accepted Answer

Call that number is $n$

Dividing $n$ by 4, 6, 8 leaves remainders 3, 5, 7 respectively, or (-1), (-1), (-1) respectively.

Hence $n+1$ is divisible by 4, 6, 8 or $n+1$ is divisible by $LCM(4,6,8)=24$

Thus $n+1=24k \implies n=24k - 1$

Since $n$ has 4 digits,we have $n \leq 9,999 \implies 24k - 1 \leq 9,999 \implies k \leq 416$

Hence $k=416 \implies n=9983$

The answer is D

ScottTargetTestPrep · Answer

The best way to solve this problem is to consider the given answer choices and use the divisibility rules for 4, 6, and 8.

Recall that if a number is divisible by 4, its last two digits are divisible by 4. Using this rule, we can eliminate choice A, since 72 is divisible by 4 (which means it won’t leave a remainder of 3). We can also eliminate choice C for the same reason. For choice B, since 96 is divisible by 4, 97 will leave a remainder of 1 when it’s divided by 4; thus, we can eliminate choice B. Choices D and E remain.

Let’s consider choice D. Since 80 is divisible by 4, 83 will indeed leave a remainder of 3 when it’s divided by 4. Now let’s use the divisibility rule for 8 that says if a number is divisible by 8, its last three digits are divisible by 8. So we we can just divide 983 by 8 and see what the remainder is:

983/8 = 122 R 7

So, it does leave a remainder 7 when it’s divided by 8. Instead of using the divisibility rule for 6, let’s just divide 9983 by 6 and see what the remainder is:

9983/6 = 1,663 R 5

The remainder is indeed 5 when 9983 is divided by 6. Thus, choice D is the correct answer.

Answer: D

mikemcgarry · Answer

Dear Supermaverick , I'm happy to respond. :-)

This is a tough question. I'm not sure that this would be on the GMAT, because of the nature of the calculations it requires. Here's how I would think about it. Let p & q be some positive integers. We know that the number is of the form (8p + 7). If that's true, then the remainder when we divide by 4 will be 3; since that requirement is automatically included, we can ignore it. We know that the number is of the form (6q + 5). First, we can find the smallest positive integer that satisfies these conditions. Go through the numbers that satisfy the first requirement and see whether it satisfies the second. p = 1 so (8p + 7) = 15. Subtract 5, and we get 10, which is not a multiple of 6. p = 2 so (8p + 7) = 23. Subtract 5, and we get 18, which is a multiple of 6! That's our number! The number 23 is of this form. All the properties of 6 and 8 would repeat at every 24, which is the LCM of 6 and 8. Any number of the form (23 + 24n) would satisfy these conditions. (Let n be some positive integer.) Here it helps to know about the rules of multiples . 24*4 = 96 is a multiple of 24 24*400 = 9600 is a multiple of 24 9600 + 240 = 9840 is a multiple of 24 9840 + 96 = 9936 is a multiple of 24 9936 + 24 = 9960 is a multiple of 24 That's close enough to 10,000 Now, add 23 to this multiple of 24. 9960 + 23 = 9983 OA = (D) Great question. Does all this make sense? Mike :-)

Shiv2016 · Answer

Hey  mike

I am not very sure how correct my approach for this question is but this is the way I solved it:

As we are given that we have to find a FOUR digit number which gives 3, 5, and 7 as remainder when divided by 4,6, and 8 respectively.

So we can write the number as:

1000a+100b+10c+d

Case 1: Division by 4

1000 a and 100 b are divisible by 4 and thus leave no remainder.

The expression (10c+d) leaves a remainder 3 when divided by 4. 
We can notice one thing here that 10c will end with 0 and thus d has to be 3.

That means 10c+d is 3 more than the multiples of 4 (I confirmed it by plugging some nos. here)

Therefore d=3 i.e. the last term is 3.

Thus we can eliminate options A, B, and C right away.

We are now left with two options.

We know both these nos. leave a remainder 3 when divided by 4. We have to just check their divisibility by 6 and 8.

Therefore the answer is D.

Is this approach correct?

Adityagmatclub · Answer

I simply divide the answer options by 4,6,8 using divisibility rules for these numbers. I solved the question inside 1.00 minutes. Answer option is D Also, I pick the options from last. :-)

gracie · Answer

lcm of 4,6,8=24
9999/24=416.6
416*24=9984
9984-1=9983
D

Mansoor50 · Answer

excellent application of divisibility rules to eliminate answer choices

even though this question can be best done by lcm(4,6,8)k + 23, the use of div rules allows for quick eliminations is pretty cool and for someone who is relearning number properties, the div rules suddenly are brought to the forefront of how to use them

thanks!!

SchruteDwight · Answer

We know that any number divided by 3 leaves a remainder that is the same as when we divide the sum of the digits by 3.

For instance, 73 leaves a remainder of  (7+3) \bmod 3 = 1

If we are looking for  x  then:  4k+3, 6b+5 \implies x=24m+11 \implies x=3k+2

Only A and D are left,

D \bmod 8 = 7 , hence D

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What is the largest four digit number when it is divided by 4, 6 and 8

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