Assuming you mean the positive multiples of 4 less than 117:
117 ÷ 4 = 291/4 → the required multiples are 1 x 4, 2 x 4, ..., 29 x 4:
4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 104, 108, 112, 116
Otherwise all the negative multiples of 4 are included (-4, -8, -12, ...) and there are infinitely many of them.
There are 7 such numbers: 4, 8, 12, 16, 20, 24, 28.
24, 28, 32, 36, 40, 44, 48.
To find numbers that are multiples of both 3 and 5, we need to find the numbers that are common multiples of both 3 and 5. These are numbers that are divisible by the least common multiple of 3 and 5, which is 15. The first four numbers less than 70 that are multiples of both 3 and 5 are 15, 30, 45, and 60.
12 is the only number which satisfies this. Multiples of 4 are: 4, 8, 12, 16, . . . .
2, 4, 6, 8, 10, 12, 14, 16, and 18.
117 < -4 but this is wrong because -4 is less than 117
No. All of the multiples of 4 are even.
They are 4 and 8
4,8,12,16,20,24,28,32,36,40,44,48,52,56,60,64,68,72,76,80,84,88,92,96
44 and 88 are the only common multiples of 4 and 11 that are less than 100
Anything less than 48.
-4, -8, -12 and so on.
The multiples of 4 that are less than 30 are 4, 8, 12, 16, 20, 24, 28
The following numbers are multiples of 4: 52 56 60 64 68
4, 8, 12, 16, 20.
The multiples of 4 with products equal or less than 20 are: 1 x 4 = 4 2 x 4 = 8 3 x 4 = 12 4 x 4 = 16 5 x 4 = 20
Their sum is 1200.