Find the least number which when divided by 6, 7, 8, 9 and 12 leaves the remainder 1 in each case.

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Solution

Dear student,

$G i v e n n u m b e r s, 6, 7, 8, 9 a n d 12 P r i m e f a c t o r i z a t i o n o f, 6 = 2 \times 3 7 = 1 \times 7 8 = 2^{3} 9 = 3^{2} 12 = 2^{2} \times 3 L C M = p r o d u c t o f e a c h p r i m e f a c t o r o f h i g h e s t p o w e r L C M = 2^{3} \times 3^{2} \times 7 = 504 504 i s t h e l e a s t n u m b e r e x a c t l y d i v i s i b l e b y 6, 7, 8, 9 a n d 12 . H e n c e, l e a s t n u m b e r w h i c h w h e n d i v i d e d b y 6, 7, 8, 9 a n d 12 l e a v e s r e m a i n d e r 1 i n e a c h c a s e = 504 + 1 = 505$

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