A 99-digit number consists of only the digits 4 and 8. The number is

Question

A 99-digit number consists of only the digits 4 and 8. The number is also divisible by 72. What is the minimum value of the sum of the digits of the number?

(A) 360

(B) 369

(C) 387

(D) 412

(E) 432

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Anurocks1233 · Answer

So for a number to be divisible by 72 it should be divisible by a 8 and 9.
For it to be divisible by 9 sum of the number of digits should be divisible by 9 
So 99*4 is divisible by but not by 8
So the the number 98 times 4 and 8 would be divisible by 8 but not by 9.
Next number 95 times 4 and four 8.
So this is divisible by both 
Hence

412

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JerryAtDreamScore · Answer

Trick Method:

Note the smallest possible answer is 396 anyways with 4*99 = 396, so ABC can already be eliminated, allowing us to choose between D and E. Choice D is not divisible by 9, thus we can choose E solely using the process of elimination.

Proper Proof:

For the number to be divisible by 72, we need the number to be divisible by 8 and 9.

Let us start with forcing the number to be divisible by 8, we need the last three digits to be a multiple of 8. 444 is the smallest number we can try, but it is not divisible by 8, so we can try 448 next and that is divisible by 8. Thus the last three digits have to be 448.

Next, let us make the rest of the digits 4, so the sum of digits would be  99*4 + 4 = 400 . We need this sum to be divisible by 9, this can be done by having 90 of 4's and 9 of 8's. Thus we need 8 more 8's, and that would bring the sum up to  400 + 8*4 = 432.

Therefore, the total is 432.

Answer: E

Kinshook · Answer

A 99-digit number consists of only the digits 4 and 8. The number is also divisible by 72. What is the minimum value of the sum of the digits of the number?

72 = 2^3*3^2

The number must be divisible by 8 & 9 both.

Divisibility rule of 8
Last 3 digits of the number must be divisible by 8.

Let us take 444; 444 divided by 8; Quotient = 55; Remainder = 4
448: 448 divided by 8; Quotient = 56; Remainder = 0

Divisibility rule 9
Sum of digits must be divisible by 9.

Let us take 4444.......448 
Sum of digits = 98*4 + 8 = 400; 400 divisible by 9; Quotient = 44; Remainder = 4
We have to add 9x + 5 to sum of digits without changing last 3 digits = {5,14,23,32,....}
5, 14 & 23 cannot be made using digits 4, therefore 32 need to be added

32 = 4*8

Sum of digits (minimum) = 400 + 32 = 432

IMO E

snarayan · Answer

possible digit values: 4/8
min value = 99*4= 396
thus possible answers are opt d/e

say # 4s= a and #8's =(99-a)
test opt d: sum of digits: 412
then 4a+(99-a)8=412-> we get a=(380/4) in admissible as 'a' has to be integer (i.e non decimal)

test opt e: sum of digits = 432
then 4a+(99-a)8=432-> yields a= 90 (# 4s= 90) and therefore #8s= 9
ans: opt e

metelmannmoritz · Answer

Must be divisible by 8 and by 9

Thus: x / (2^3 * 3^2)

We can see 2^3 = 8 and there must be at least nine (3^2) eight's in x. Since there must be nine eights, in order for x to be minimized, the remaining 90 digits must be 4 (not 8).

Thus there are nine 8's and 90 4's

9*8 + 90 * 4 = 432

A 99-digit number consists of only the digits 4 and 8. The number is

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A 99-digit number consists of only the digits 4 and 8. The number is

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