How many three-digit multiples of 18 have the sum of the digits also equal 18?
Answers
To find:
How many three-digit multiples of 18 have the sum of the digits also equal 18
Step-by-step explanation:
The three-digit multiples of 18 are 108, 126, 144, 162, 180, 198, 216, 234, 252, 270, 288, 306, 324, 342, 360, 378, 396, 414, 432, 450, 468, 486, 504, 522, 540, 558, 576, 594, 612, 630, 648, 666, 684, 702, 720, 738, 756, 774, 792, 810, 828, 846, 864, 882, 900, 918, 936, 954, 972 and 990.
Of these, the numbers whose sum of the digits also equal 18, are 198, 288, 378, 396, 468, 486, 558, 576, 594, 648, 666, 684, 738, 756, 774, 792, 828, 846, 864, 882, 918, 936, 954, 972 and 990, which are 25 in number.
Check for a few numbers:
198 = 1 + 9 + 8 = 18
288 = 2 + 8 + 8 = 18
378 = 3 + 7 + 8 = 18
396 = 3 + 9 + 6 = 18
468 = 4 + 6 + 8 = 18 and so on.
Answer:
25 three-digit multiples of 18 have the sum of the digits also equal 18.
Answer:
25
Step-by-step explanation:
This explanation has 4 steps
18 is an even number which means the last digit of its multiples would be 0,2,4,6,8.
H T U
a) _ _ 0
b) _ _ 2
c) _ _4
d) _ _ 6
e) _ _8
If the sum of the three digits is 18, then by subtracting the last digit from 18 we get the sum of the first two digits.
a) _ _ 0 = 18 (18-0)
b) _ _ 2 = 16 (18-2)
c) _ _4 = 14 (18-4)
d) _ _ 6 = 12 (18-6)
e) _ _8 = 10 (18-10)
By figuring out the possible combinations of the sums of two digits, we will get the multiples of 18.
a) _ _ 0 = 18 => (9+9) = 1
b) _ _ 2 = 16 => (9+7), (8+8) = 3
c) _ _4 = 14 => (9+5), (8+6), (7+7) = 5
d) _ _ 6 = 12 => (9+3), (8+4), (7+5), (6+6) = 7
e) _ _8 = 10 => (9+1), (8+2), (7+3), (6+4), (5+5) = 9
(Pairs with different digits would be counted twice since these digits would result in two multiples of 18. e.g.: (9+7) would become 792 and 972)
Lastly, we shall add up all the numbers that are underlined. These number show us the count of our possible combinations for the multiples
1+3+5+7+9=25
25 three-digit multiples of 18 have the sum of the digits which also equal 18.