7. Find the smallest positive integer x such that when x is divided by 3 the rem

Question

7. Find the smallest positive integer x such that when x is divided by 3 the remainder is 2, when it is divided by 7 the remainder is 4, and when it is divided by 10 the remainder is 6
7. Find the smallest positive integer x such that when x is divided by 3 the remainder is 2, when it is divided by 7 the remainder is 4, and when it is divided by 10 the remainder is 6
7. Find the smallest positive integer x such that when x is divided by 3 the remainder is 2, when it is divided by 7 the remainder is 4, and when it is divided by 10 the remainder is 6

Explanation / Answer

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Given That positive integer when divided by 10 leaves a remainder of 6

=> The number shall end with 6 i.e 6,16 ,26,36 .......

Now since this same number when divided by 3 leaves remainder as 2

=> The smallest number divisible by 3 with remainder as 2 should be

6/3 = 0 remainder X

16/3 =1 remainder X

26 /3 =2 remainder (Correct)

which shall repeat itself after ever 30 integers

Now the thrid requirement is that the number on division by 7 shall leave a remainder of 4

we shall be checking integers like 26,56,86 ....

So the smallest integer is 206

Hence solved.

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