The number 523abc is divisible by 7,8,9. Then what is the

I also choose D.
a, b, c cannot be 0.
The number is divisible by 9 =>sum of 5+2+3+a+b+c must be divisible by 9, so a+b+c can be 8, 17 or 26 (as maximum sum of 3 1-digit number is 27).
It is also divisible by 8 => It will be divisible by 4 and 2 => bc must be divisible by 4 and c must be even. so c just can be 2 or 6.
It is also divisible by 7.
So abc must be 656, its product is 180.

I broke the answer choices into their prime factors, then tested the different 3 digit combos ending in an even number for which one would make 5+2+3+a+b+c be divisible by 9. The only number with prime factors that does that is 180. Hence, the answer is D.

I thought that for a number to be divisible by 8, its last three digits had to be divisible by 8, any clue on why this is not working in answer choice D?

Cheers!
J

The options give us the value of a*b*c, not abc. If we get abc as 656, it is divisible by 8. The options give us a*b*c = 6*5*6 = 180. The product may or may not be divisible by 8.

Thanks Karishma, anyways what I tried here was prime factorization to see which answer choice could be multiples of 9, that is having the whole sum of digits to be a multiple of 9 and only C and D worked. Now, it seems a bit complicated to decide between the two. I have both even too. Any clues on how to proceed from here? I still haven't used the fact that they need to be divisible by 7 and 8 but I'm not sure how to approach these tests quickly

Kudos for a good reply!
Cheers
J

523abc is divisible by 9 which means 5+2+3+a+b+c is divisible by 9 i.e. 1+a+b+c is divisible by 9.

When you prime factorize each option, you are left only with (D)

A. \(504 = 2^3 * 3^2 * 7\)
Since a, b and c must be single digit numbers, the only way you can assign values to them is 8, 9 and 7. 1+8+9+7 = 25 which is not divisible by 9.

B. \(532 = 2^2*7*19\)
19 cannot be assigned to any of a, b and c and hence this product is not possible.

C. \(210 = 2^2 * 3 * 5 * 7\)
You cannot combine them in any way such that you get three single digit numbers. Note that if 5 and 7 are multiplied by even 2, they will become double digit hence two of a, b and c must take values of 5 and 7. The third cannot be the leftover 12.

D. \(180 = 2^2 * 3^2 * 5\)
There are two options: 4, 9, 5 and 6, 6, 5.
6, 6, 5 satisfies since 1+6+6+5 = 18 divisible by 9

E. \(280 = 2^3 * 5 * 7\)
The digits must be 8, 5 and 7. But 1+8+5+7 is not divisible by 9.

Answer (D)

I have a doubt Bunuel. Wouldn't that be equal to aba? because i tried solving it backwards using options. The only way option D could've factorized into single digits was 4,5,9 and for this case the singular digits would add up to 5+2+3+4+5+9 = 28 which would make it indivisible by 9. i understand it could have been 6,6,5 too but wouldn't that make it aba instead of abc?

I think you assume that a, b and c must be distinct digits. But this is not true: unless it is explicitly stated otherwise, different variables CAN represent the same number.

Hope it's clear.

I find this method to be very time consuming.

LCM of 7,8 & 9 = 7 * 8 * 9 = 504

Whatever the number 523abc should be divisible by 504

We know that 504000 is divisible by 504

Adding multiples of 504 to 504000 to reach 523abc

We get two numbers:

523152 & 523656 (both are divisible by 504)

Product of abc = 10 OR 180

Option D fits in

Answer = 180 = D

523abc is divisible by 504.

To find abc first lets divide 523000 by 504

523000/504 = 137 and remainder is 352. -------------(1)

352 + 152 = 504
i.e. when 152 is added with the remainder 352 of (1) then in (1) quotient = 138 and remainder will be 0

523152 / 504 = 138 and remainder is 0
As per the question "abc" is a*b*c = 180
but here 1*5*2=10

523656/504 = 139 and remainder is 0
a*b*c= 6*5*6 = 180

Regards,
Swami.

What ?
Okay so you are telling me to find this hepttyy looking divisibility computation..
Looks Wrong to me..
answer is not doubt right
but the method is tooooooooooo long
P.S => add a few more ooooo in tooooo

f(523abc)=7,8,9;

number/9: sum of digits must be a multiple of 9

number/8: last digit must be even {2,4,6,8}, but not {0} because if c=0 then abc=0.

0≤a,b,c≤9 (digits)
0≤a+b+c≤27=(9*3)

5+2+3=10 and 10+a+b+c=multiple(9), so a+b+c={8,17,26}=m(9)-1

A. f(504)=126*4=63*8=7*8*9; sum=24
B. f(532)=4*133=4*7*19; no digit can be greater than 9
C. f(210)=21*10=7*3*2*5=7*6*5; sum=18
D. f(180)=18*10=9*4*5=6*6*5; sum={17}=m(9)-1
E. f(280)=28*10=7*4*2*5=7*8*5; sum=20

Ans (D)

You divided 523000 by 504 with no calculator to get the 352 remainder?

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