Question 1170032
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(a) multiple of 216?<br>
216 = (2^3)(3^3)<br>
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.<br>
ANSWER: Yes, P can be a multiple of 216.<br>
(b) multiple of 2000?<br>
2000 = (2^4)(5^3)<br>
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.<br>
ANSWER: No, P can't be a multiple of 2000.<br>
(c) number of values of P that are multiples of 128 but not of 1024?<br>
128 = 2^7; 1024 = 2^10<br>
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.<br>
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.<br>
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).<br>
ANSWER: Only one value of P divisible by 128; and it is not divisible by 1024.<br>
(d) number of ORDERED quadruples for which P is 98 less than a multiple of 100 (2, 102, 202, 302, ...)<br>
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.<br>
2 = 1*1*1*2<br>
Since this question asks for the number of ORDERED quadruples, the answer is 4 (the single 2 can be any of the four numbers).<br>
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<br>