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sourangshu maiti

· started a discussion

· 1 Months ago

do not understand (Question: 20368.0 )

how we know the valu of "k" if please tell

Question:
The least multiple of 13, which on dividing by 4, 5, 6, 7 and 8 leaves remainder 2 in each case, is?
Options:
A) 2,520
B) 842
C) 2,522
D) 840
E) None of these.
Solution:
Here,

\(\begin{array}{r|r r r} 2 & 4, & 5, & 6, &7, &8 \\ \hline 2 & 2, & 5, & 3,&7,&4 \\ \hline 2 & 1, & 5, & 3,&7,&2 \\ \hline 3 & 1, & 5, & 3,&7,&1 \\ \hline 5 & 1, & 5, & 1,&7,&1 \\ \hline 7 &1,&1,&1,&7,&1\\ \hline &1,&1,&1,&1,&1 \end{array}\)

therefore LCM (4,5,6,7,8) = 2x2x2x3x5x7=840

But 2 is the remainder in each case

therefore,  Required number = 840K + 2

This should be multiple of 13 

for K = 3

The number is = 840 x 3 + 2 = 2522

Which is multiple of 13

kanhaiya jha

· commented

· 1 Months ago

Take first 5 natural numbers of k and put in 840k+2 and check the condition

sourangshu maiti

· commented

· 1 Months ago

detail descrup tion

kanhaiya jha

· commented

· 1 Months ago

LCM of 4, 5, 6, 7 and 8 = 840.

Suppose, require number be 840K + 2 which is multiple of 13.

Least value of K for which (840K + 2) is divisible by 13 is K = 3

∴ Required Number = 840 * 3 + 2 = 2520 + 2 = 2522.
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