A number theory problem by shreyas shastry

Number Theory Level 2

Find the least positive number which when divided by 2,3,4,5,6, leaves remainder of 1 in each case but when divided by 7 leaves no remainder. find a smallest such number.


The answer is 301.

Shreyas Shastry by shreyas shastry , 22, India

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2 solutions

Aashish Patel
Apr 17, 2014

Firstly we have to find the LCM of 2,3,4,5,6=60.So if we add 1 to 60 then the number will leave a remainder every time.

61/2=Remainder 1.

61/3=Remander 1.

61/4=Remainder 1.

61/5=Remainder 1.

61/6=Remainder 1.

But 61 is not divisible by 7 ,so:

The next numbers are-

120+1,

180+1,

240+1,

300+1.

Hence at last 301 is divisible by 7.So it is the answer.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

shouldn't it be 91?

Anjaneya Tiwari - 7 years, 1 month ago

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Sorry my bad

Anjaneya Tiwari - 7 years, 1 month ago

301

Shiwang Srivastava - 7 years, 1 month ago
Sunil Pradhan
Apr 16, 2014

least number which when divided by 2,3,4,5,6, leaves remainder of 1 is LCM of (2, 3, 4, 5, 6) + 1 = 61

But 61 is not divisible by 7 next numbers are (120 + 1), (180 + 1), (240 + 1), (300 + 1) and so on out of these least number divisible by 7 is 300 + 1 = 301

1 Helpful 0 Interesting 0 Brilliant 0 Confused

60k+1 and it is equivalent to 4k+1 congruent mod 7, so if it is divisible by 7 then 4k+1=0(mod 7), hence the least such k is 5 and the required number is 301

Subhajit Jana - 7 years, 1 month ago

more precisely 4k=-1(mod 7)=> 4k=6(mod 7)=> 2k=3(mod 7)=> 2k=10(mod 7)=>k=5(mod 7) as 2 is prime to 7.........hence the required k=5

Subhajit Jana - 7 years, 1 month ago

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