989. If there is a remainder of 2 when divided by 3, the number is one less than a multiple of 3. If there is a remainder of 4 when divided by 5, the number is one less than a multiple of 5. Thus the number required is one less than a multiple of the lowest common multiple of 3 and 5 (that is 15). So what is needed is an even multiple of 15 less than or equal to 1000: 1000 ÷ 15 = 662/3 Thus the highest even multiple of 15 not greater than 1000 is 66 x 15 = 990, and the required number is 989.
3
Given any number, there is an even number that exists greater than it. That even number is a product: of 2 and some number. Therefore, the number that you started with is less than the product of a pair of numbers.
and even # less than 50 is any number smaller than 50 ending in the number 2,4,6,8,0
27
150
989. If there is a remainder of 2 when divided by 3, the number is one less than a multiple of 3. If there is a remainder of 4 when divided by 5, the number is one less than a multiple of 5. Thus the number required is one less than a multiple of the lowest common multiple of 3 and 5 (that is 15). So what is needed is an even multiple of 15 less than or equal to 1000: 1000 ÷ 15 = 662/3 Thus the highest even multiple of 15 not greater than 1000 is 66 x 15 = 990, and the required number is 989.
Every prime number except 2 must be odd. This is because if it were even it would be divisible by 2 and so wouldn't be a prime number. The next point to note is that all odd numbers are one more or one less than a multiple of 4. Because 2 more than a multiple of 4 is even, and 3 more than a multiple of 4 is the same as 1 less than the multiple of 4 above. Thus every prime number except 2 must be 1 more or 1 less than a multiple of 4.
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.
There are 49 even numbers less than 100, but only the number 2 is an even prime number less than 100.
4 x 5
An even number less than 57.
3
Given any number, there is an even number that exists greater than it. That even number is a product: of 2 and some number. Therefore, the number that you started with is less than the product of a pair of numbers.
and even # less than 50 is any number smaller than 50 ending in the number 2,4,6,8,0
84
Five of them.