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For any integer n, f(n) is the greatest multiple of 10 less than or equal to n + 5. If a, b, c, and d are positive integers such that abcd = 360, what is the greatest possible value of f(a) + f(b) + f(c) + f(d) − (a + b + c + d)?
A) 7
B) 9
C) 10
D) 11
E) 13
A) 7
B) 9
C) 10
D) 11
E) 13
04 Apr 2026, 05:59
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Think about what the function is doing. You take n, add 5, and then round down to the nearest multiple of 10. So the only thing that really matters is the units digit of n.
If n ends in 5, then n + 5 lands exactly on a multiple of 10, so you gain the most: +5.
If it ends in 6, you still jump up a lot: +4.
Then 7 gives +3, 8 gives +2, 9 gives +1.
If it ends in 1 to 4, you actually lose value. So those are bad.
So the goal is simple: make the numbers end in 5 or 6 as much as possible.
Now look at the product: abcd = 360 = \(2^3 × 3^2 × 5\).
There is only one factor of 5, so only one number can end in 5. The others have to be made from 2s and 3s.
If n ends in 5, then n + 5 lands exactly on a multiple of 10, so you gain the most: +5.
If it ends in 6, you still jump up a lot: +4.
Then 7 gives +3, 8 gives +2, 9 gives +1.
If it ends in 1 to 4, you actually lose value. So those are bad.
So the goal is simple: make the numbers end in 5 or 6 as much as possible.
Now look at the product: abcd = 360 = \(2^3 × 3^2 × 5\).
There is only one factor of 5, so only one number can end in 5. The others have to be made from 2s and 3s.
Updated on: 04 Apr 2026, 13:32
Originally posted by GMATcracker101 on 04 Apr 2026, 06:26.
Last edited by GMATcracker101 on 04 Apr 2026, 13:32, edited 1 time in total.
Last edited by GMATcracker101 on 04 Apr 2026, 13:32, edited 1 time in total.
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Yes, I get that.
if we take - 360 = a*b*c*d = 9*8*5*1 or 6*6*5*2
f(n) = greatest multiple of 10 less than or equal to n+5
Accordingly -
Case I - 360 = a*b*c*d = 9*8*5*1
f(a) = 10 as 9 + 5 = 14 and the greatest multiple of 10 less than or equal to 14 is 10.
f(b) = 10 as 8 + 5 = 13 and the greatest multiple of 10 less than or equal to 13 is 10.
f(c) = 10 as 5 + 5 = 10 and the greatest multiple of 10 less than or equal to 10 is 10.
f(d) = 0 as 1 + 5 = 6 and the greatest multiple of 10 less than or equal to 6 is 0.
f(a+b+c+d) = 20 as 23 + 5 = 28, the greatest multiple of 10 less than or equal to 28 is 20.
Hence, 30 - 20 = 10.
Case II - 360 = a*b*c*d = 6*6*5*2
f(a) = 10 as 6 + 5 = 11 and the greatest multiple of 10 less than or equal to 11 is 10.
f(b) = 10 as 6 + 5 = 11 and the greatest multiple of 10 less than or equal to 11 is 10.
f(c) = 10 as 5 + 5 = 10 and the greatest multiple of 10 less than or equal to 10 is 10.
f(d) = 0 as 2 + 5 = 7 and the greatest multiple of 10 less than or equal to 7 is 0.
f(a+b+c+d) = 20 as 19 + 5 = 24 and the greatest multiple of 10 less than or equal to 24 is 10.
Hence, 30 - 20 = 10
Kindly explain how the answer is 11.
if we take - 360 = a*b*c*d = 9*8*5*1 or 6*6*5*2
f(n) = greatest multiple of 10 less than or equal to n+5
Accordingly -
Case I - 360 = a*b*c*d = 9*8*5*1
f(a) = 10 as 9 + 5 = 14 and the greatest multiple of 10 less than or equal to 14 is 10.
f(b) = 10 as 8 + 5 = 13 and the greatest multiple of 10 less than or equal to 13 is 10.
f(c) = 10 as 5 + 5 = 10 and the greatest multiple of 10 less than or equal to 10 is 10.
f(d) = 0 as 1 + 5 = 6 and the greatest multiple of 10 less than or equal to 6 is 0.
f(a+b+c+d) = 20 as 23 + 5 = 28, the greatest multiple of 10 less than or equal to 28 is 20.
Hence, 30 - 20 = 10.
Case II - 360 = a*b*c*d = 6*6*5*2
f(a) = 10 as 6 + 5 = 11 and the greatest multiple of 10 less than or equal to 11 is 10.
f(b) = 10 as 6 + 5 = 11 and the greatest multiple of 10 less than or equal to 11 is 10.
f(c) = 10 as 5 + 5 = 10 and the greatest multiple of 10 less than or equal to 10 is 10.
f(d) = 0 as 2 + 5 = 7 and the greatest multiple of 10 less than or equal to 7 is 0.
f(a+b+c+d) = 20 as 19 + 5 = 24 and the greatest multiple of 10 less than or equal to 24 is 10.
Hence, 30 - 20 = 10
Kindly explain how the answer is 11.
