JUST A MODULO PROBLEM!!(2)

Number Theory Level 3

A number X leaves no remainder when divided by 7, 5 when divided by 6, 4 when divided by 5, 3 when divided by 4, 2 when divided by 3, 1 when divided by 2? Find the least value of X?


The answer is 119.

Jaiveer Shekhawat by jaiveer shekhawat , 22, India

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3 solutions

Sunil Pradhan
Oct 8, 2014

No. ÷ 7 remainder 0, No. ÷ 6 remainder 5, No. ÷ 5 remainder 4

No. ÷ 4 remainder 3, No. ÷ 2 remainder 1

Every time except 7 the difference in divisor and remainder = 1

(6 – 5), (5 – 4), etc = 1

in such case least number = LCM of divisors – difference

= (LCM of 2, 3, 4,5, 6) – 1 = 60 – 1 = 59

But 59 is not divisible by 7 so next number is 60 × 2 – 1 = 119

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James Moors
Oct 1, 2014

X = 7 g = 6 f − 1 = 5 e − 1 = 4 d − 1 = 3 c − 1 = 2 b − 1 X=7g=6f-1=5e-1=4d-1=3c-1=2b-1 .

L C M ( 6 , 5 , 4 , 3 , 2 ) = 60 LCM(6,5,4,3,2) = 60 so X = 7 g = 60 a − 1 X=7g=60a-1 .

60 ≡ 4 ( m o d 7 ) ⟹ 120 − 1 ≡ 0 ( m o d 7 ) 60 \equiv 4 (mod 7) \implies 120-1 \equiv 0 (mod 7)

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the number can be written as 7k/3l+2/5m+4/2k+1/6s+5. Thus we see last digit has to be 9. We see all multiple of 7 with last digit 9. tHE Least one satisfies this is 119

@jaiveer shekhawat - how did u do it?

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well I first took the lcm of 1 to 6 which is 60 and then checked that:

60 = 4(mod 7)

thus multiplied it by 2 on both sides and i got

120=1(mod7)

=119=0(mod7)

that's it!!

jaiveer shekhawat - 6 years, 8 months ago

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