Q: Must choose a number between 67 and 113 that is a multiple of 58 and 10?

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It is: 78 times 6 = 468

80

80 adn 100

Each of these is correct: 54, 63, 72, 81 and 90.

60 & 80

84 Because 6 is a multiple of 3, 3 doesn't really count. 6 times 7 is 42. a multiple of 42 that is between 61 and 107 is 84. That is why 84 is the answer.

A multiple-choice question is one for which the respondent must choose from a list of predetermined responses.

A multiple of 356 would be a number than 356 can divide into. So there is no number between 67 and 113 that can be divided by 356. The number 89 is between is between 67 and 113 and it can be divided into 356 evenly, so it is a factor, but not a multiple. 356 is a multiple of 89.

To be a multiple of 4, 10 and 20 the number must be a multiple of their lowest common multiple. lcm(4, 10, 20) = 20 What is required is a multiple of 20 between 61 and 107. The numbers 80 and 100 both satisfy the problem of being between 61 and 107 and being a multiple of 4, 10 and 20.

If 245 is a factor, and it is between 200 and 300, then the number must be 245 - provided it is a multiple of 7, which it is.

yes exept for the numbers 3,9,15,21,27,33 etc

No number that is a multiple of 3, can be a prime number. A prime number must only be divisible by itself and 1. It cannot be divisible by any other number. Therefore if it is a multiple of 3, then it must be divisible by 3 and hence, not a prime number.