Every number has 1 as a multiple. 1*100 = 100 1*7 = 7 As 1 is an odd number there are no numbers with only even multiples.
Any number that is divisible by 2 and is a multiple of that particular number is called an even multiple. Ex: The even multiples of 7 are: 14, 28 etc
A multiple is the result of multiplying a number by an integer. 7: 7, 14, 21, 28
The smallest multiple of 7 that's larger than 1 is the first multiple . . . 7.The largest multiple of 7 that's smaller than 200 is the 28th multiple . . . 196 .So there are 28 of them in that range.
70
The three digit number satisfying the requirements is 112. The three digit number must be greater than or equal to 100 and less than 140. To have 7 as a factor it must be a multiple of 7 To be even, it must be an even multiple of 7. The first even multiple of 7 greater than or equal to 100 is: 100 ÷ 7 = 14 r 2 → first even multiple is 16 × 7 = 112 The last even multiples of 7 less than 140 is: 140 ÷ 7 = 20 → last even multiple is 18 × 7 = 126 The sum of its digits must be less than 9: 112 → 1 + 1 + 2 = 4 126 → 1 + 2 + 6 = 9 → only 112 fits all the criteria.
70
It is 70
Every number has 1 as a multiple. 1*100 = 100 1*7 = 7 As 1 is an odd number there are no numbers with only even multiples.
If you take three consecutive odd (or three consecutive even) numbers, one of the three will always be a multiple of 3.If you take three consecutive odd (or three consecutive even) numbers, one of the three will always be a multiple of 3.If you take three consecutive odd (or three consecutive even) numbers, one of the three will always be a multiple of 3.If you take three consecutive odd (or three consecutive even) numbers, one of the three will always be a multiple of 3.
35 5X7 = 35
105 is a multiple of 7
There is no solution. The sum of multiple even integers will always be even, and -7 is odd.
No, 92 is not an even multiple of 7 (92/7= 13.1428571).
It could be: 4*7 = 28
There are no three consecutive even integers that add up to 176.Let the three consecutive even integers be (2n-2), 2n & (2n+2). Then their sum is:(2n-2) + 2n + (2n+2) = 6n⇒ the total must be a multiple of 6.To be a multiple of 6 the number must be an even multiple of 3; for 176:It is even;1 + 7 + 6 = 14; 1 + 4 = 5 which has a remainder of 2 when divided by 3, so 176 is not a multiple of 3.Thus 176 is not a multiple of 6 and so cannot be the sum of three consecutive even integers.
There are no first 5 multiple. Multiple is singular.