23, 58, 93, 128, 163, 198, ...
3
No. Add the digits of the dividend and if that is divisible by 3 then the original number is divisible by 3; if not, its remainder when divided by 3 gives the remainder when the original number is divided by 3: 1 + 2 + 1 = 4 which gives a remainder of 1 when divided by 3, so 121 divided by 3 gives a remainder of 1. (121 = 40 x 3 + 1)
MOD (or modulus) gives the remainder when the first number is divided by the second. The remainder of 16 divided by 2 is 0, so 16 MOD 2 = 0.
It is an Odd number. Even numbers are those which have no remainder when divided by 2.
236 divided by 5 is 47 with remainder 1
3
23 is an odd number between 10 and 30 that gives a remainder of 2 when divided by 7.
172
3 divided by 2 goes once, remainder one.
No. Add the digits of the dividend and if that is divisible by 3 then the original number is divisible by 3; if not, its remainder when divided by 3 gives the remainder when the original number is divided by 3: 1 + 2 + 1 = 4 which gives a remainder of 1 when divided by 3, so 121 divided by 3 gives a remainder of 1. (121 = 40 x 3 + 1)
I think you're wanting a number of two digits, one of which is 3, that when divided by 7 gives a quotient and a remainder of 1 and when that quotient is divided by 2 it gives a remainder of 1: Answer: 36 36 ÷ 7 = 5 r 1 5 ÷ 2 = 2 r 1 If you want the number to be such that if it is divided by 7 the remainder is 1 and if it is divided by 2 the remainder is 1, then: Answer: 43 43 ÷ 7 = 6 r 1 43 ÷ 2 = 21 r 1
3 divided by 2 gives 1r1
59
29
MOD (or modulus) gives the remainder when the first number is divided by the second. The remainder of 16 divided by 2 is 0, so 16 MOD 2 = 0.
It is an Odd number. Even numbers are those which have no remainder when divided by 2.
62 explanation is : 4X15+2=62