Problema Solution
on a stranded die with six faces each face contains a different number from 1 through 6. Jake has a nonstandard die with six faces and each face on Jake's die contains an expression with a different value from through 6. in no particular order the six expressions are a+1 ,2a-5 , 3a-10 , b+8 , 2b+5 , and 3b+10. if a and b are integers what is the value of the product a times b
Answer provided by our tutors
The sum of all six expressions must equal 1 + 2 + 3 + 4 + 5 + 6 = 21, so:
(a + 1) + (2a − 5) + (3a − 10) + (b + 8) + (2b + 5) + (3b + 10) = 21
6a + 6b + 9 = 21
6a + 6b = 12
a + b = 2
Given the expressions containing b, we see that b must be negative in order to yield an integer result from 1 to 6, while a must be positive.
For the expression 3b + 10 to have a value from 1 to 6, b must have a value from −1 to 3, inclusive.
If b = −1, then b + 8 = 7, which is greater than 6.
If b = −2, then 3b + 10 = 4, 2b + 5 = 1, b + 8 = 6. So far, these sides have values that add up to 4 + 1 + 6 = 11. And for the sum of the values on the remaining three sides we’d have:
(a + 1) + (2a − 5) + (3a − 10) = 21 − 11
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a = 4
But we still need to make sure these three sides have the correct values.
If a = 4, then a + 1 = 5, 2a − 5 = 3, 3a − 10 = 2.
When a = 4 and b = −2, we have the values 1 through 6 on the faces of the die and a × b = 4 × (−2) = −8.