SOLUTION: The integers a, b, c, and d are each equal to one of 1, 2, 3, 4, 5, 6, 7, 8, 9. (It is possible that two or more of the integers a, b, c, and d have the same value.) The integer

Question 1170032: The integers a, b, c, and d are each equal to one of 1, 2, 3, 4, 5, 6, 7, 8, 9. (It is
possible that two or more of the integers a, b, c, and d have the same value.) The
integer P equals the product of a, b, c, and d; that is, P = abcd.
(a) Determine whether or not P can be a multiple of 216.
(b) Determine whether or not P can be a multiple of 2000.
(c) Determine the number of possible different values of P that are divisible by 128
but are not divisible by 1024.
(d) Determine the number of ordered quadruples (a, b, c, d) for which P is 98 less
than a multiple of 100.
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

(a) multiple of 216?

216 = (2^3)(3^3)

In choosing 4 (not necessarily different) integers from 1 to 9, we need at least 3 factors of 2 and at least 3 factors of 3.
9*8*3 gives us 3 factors of 2 and 3 factors of 3; so (for example) we could choose those 3 numbers and any of the 9 numbers for the fourth.

ANSWER: Yes, P can be a multiple of 216.

(b) multiple of 2000?

2000 = (2^4)(5^3)

Among the integers 1 through 5, there is only one factor of 5. So we would need to choose 5 for 3 of the 4 numbers. But then the fourth number would have to contain 4 factors of 2, which it can't.

ANSWER: No, P can't be a multiple of 2000.

(c) number of values of P that are multiples of 128 but not of 1024?

128 = 2^7; 1024 = 2^10

We are looking for combinations of 4 of the integers from 1 to 9 that together contain at least 7 factors of 2 but not 10.

To get 7 factors of 2, we need to use all the even numbers, 2, 4, 6, and 8; together they have exactly 7 factors of 2.

Since we had to use 4 of the integers to get the required 7 factors of 2, there is only the one way to get a number divisible by 128: 2*4*6*8 = 384 (= 3*128).

ANSWER: Only one value of P divisible by 128; and it is not divisible by 1024.

(d) number of ORDERED quadruples for which P is 98 less than a multiple of 100 (2, 102, 202, 302, ...)

This one requires a lot of investigation, since the product can be any of a large number of numbers. However, use of a prime factorization calculator shows that the number 2 is the only number in the list that can be written as the product of 4 integers from 1 through 9.

2 = 1*1*1*2

Since this question asks for the number of ORDERED quadruples, the answer is 4 (the single 2 can be any of the four numbers).

ANSWER: 4 ordered quadruples with P equal to 98 less than a multiple of 100:
1*1*1*2
1*1*2*1
1*2*1*1
2*1*1*1

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