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Z is a two digit positive integer, Z is equal to 7 times the sum of its digits. Is Z=84?

1. Sum of its digits is 12 2. Product of its digits is 32

Let the tenths and units digits of Z be \(a\) and \(b\) respectively, so \(Z=10a+b\).

Given: \(Z=10a+b=7(a+b)\) --> \(a=2b\) --> \(Z=10a+b=20b+b=21b\), so Z could take only the following 4 values: 21, 42, 63, and 84.

Question: is \(Z=84\)?

(1) Sum of its digits is 12 --> from the possible values of Z only 84 has the sum of its digits equal to 12. Sufficient. (2) Product of its digits is 32 --> from the possible values of Z only 84 has the product of its digits equal to 32. Sufficient.

my brain is not clicking, but can someone explain how a = 2b?

i'm understanding this part: 10a + b = 7 (a+b)

Well 10a + b is just another way of writing a two digits number.

For example, if Z = 65, then Z = 65 = 10*6 + 5. where 6 = "a" and 5 = "b" taking things further, we can express any length number using the notation above.

so if Z = 12345.6 = 10000*1 + 1000*2 + 100*3 + 10*4 + 5 + 0.1*6

and since we are given the fact that Z = 7 * (sum of the digits) = 7* (a+b) we are able state that 10a + b = 7 (a+b).

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