What two digit number is les than the sum of the square of its digit by 11 and exceeds their doubled product by 5?
Answers
Step-by-step explanation:
We are told that the two-digit number exceeds doubled product of its digits by 5:
(10a+b)−2ab=5(10a+b)−2ab=5 --> 2a(5−b)−(5−b)=02a(5−b)−(5−b)=0 --> (5−b)(2a−1)=0(5−b)(2a−1)=0 --> b=5b=5 (aa cannot equal to 1/2 since it must be an integer). The only answer choice with 5 as an units digit is A.
Check 95 for the first condition (to eliminate E), which says that the two-digit number is less than the sum of the square of its digits by 11: (9^2+5^2)-95=11. So, the answer is A.
Answer: A.
There is another number satisfying both conditions:
Substitute b=5b=5 in (a2+b2)−(10a+b)=11(a2+b2)−(10a+b)=11 --> a2−10a+9=0a2−10a+9=0 --> a=9a=9 or a=1a=1. Therefore both 15 and 95 satisfy both conditions.
Hope it's clear.
Answer:
Step-by-step explanation:
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