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Q: What is the smallest number that when divided by 5 is a remainder of 2?

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It is 30.

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Smallest prime factor = smallest prime number which can be divided into the given number with no remainder 82/2=41, both 2 and 41 are prime and 2 is the smallest.

The smallest number that can be divided by 2 3 4 5 6 7 8 and always give you a remainder of 1 is 841

The answer is 1. The smallest prime is 2 and the greatest prime (of however many digits) must be odd. The remainder of ANY odd number when divided by 2 must be 1.

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169 is the smallest such number.

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5. To leave a remainder of 1 when divided by 2, the number must be odd. To leave a remainder of 2 when divided by 3, the number must also be two more than a multiple of 3. The multiples of 3 which are odd are 1 x 3, 3 x 3, 5 x 3, etc. The smallest odd multiple of three is 1 x 3 = 3 ⇒ required number is 3 + 2 = 5.

85. A number when divided by 238, leaves a remainder 79. What will be the remainder when the number is divided bv 17 ? (1) 8 (2) 9 (3) 10 @) 11

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3 divided by 2 has a remainder of 1. Which is 1 less than 2.

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It is not possible, because the number 4 is divisible by 2, and it's remainder is divisible by 2 also, so whatever number works for the "4 with a remainder of 2", will never work for "2 with a remainder of 1.

-2

16 \;p

It is an Odd number. Even numbers are those which have no remainder when divided by 2.

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