Accenture
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Numerical Ability
Number System
Q. The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
Option
A. 74
B. 94
C. 184
D. 364
Read Solution (Total 5)
-
- D. 364
LCM of 6, 9, 15 and 18 is 90.
90*1+4 = 94 not divisible by 7.
90*2+4 = 184 not divisible by 7.
90*3+4 = 274 not divisible by 7.
90*4+4 = 364 is divisible by 7.
so 364 .. option D - 13 years agoHelpfull: Yes(20) No(2)
- when divided by 7 it leaves the remainder 4 so n-4 should be the multiple of lcm(6,9,15,18)
So lcm(6, 9, 15, 18) = 2 * 3^2 * 5 = 90.
So n - 4 must be a multiple of 90.
If n - 4 = 0, then n = 4, which doesn't work because 4 isn't a multiple of 7.
If n - 4 = 90, then n = 94, which doesn't work because 94 isn't a multiple of 7.
If n - 4 = 180, then n = 184, which doesn't work because 184 isn't a multiple of 7.
If n - 4 = 270, then n = 274, which doesn't work because 274 isn't a multiple of 7.
If n - 4 = 360, then n = 364. This is a multiple of 7, and so it's the smallest number that works. - 12 years agoHelpfull: Yes(6) No(0)
- lcm of 6,9,15 and 18 is 90
least multiple of 7 and remainder 4 takes the form 90k+4 which should be divisible by 7. putting k = 4, answer is 364 - 12 years agoHelpfull: Yes(5) No(0)
- after LCM of 6,9,15,18 we get 90 but rem is 4 then number can be 90+4=94 but it is not divisible by 7 so
same 74,94,184 are not divisible by 7 ie multiple of 7 is only 364
so Option :D
- 12 years agoHelpfull: Yes(2) No(2)
- LCM OF (6,9,15,18) = 90
90K+4 SHOULD BE DIVISIBLE BY 7
FOR K=4 , 364
DIVISIBLE BY 7
- 8 years agoHelpfull: Yes(1) No(0)
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