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19 Jan 2021, 01:15
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D
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If a and b are positive integers and ab > 13, is a ≠ b?
(1) Exactly three of the numbers ab - 4, ab - 2, ab + 1 and ab + 3 have the same tens digit
(2) ab is divisible by 8, but not divisible by 16.
Are You Up For the Challenge: 700 Level Questions
(1) Exactly three of the numbers ab - 4, ab - 2, ab + 1 and ab + 3 have the same tens digit
(2) ab is divisible by 8, but not divisible by 16.
Are You Up For the Challenge: 700 Level Questions
19 Jan 2021, 02:20
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Bunuel
If a=b, then ab will be a SQUARE. So we have to check whether ab is a perfect square.
(1) Exactly three of the numbers ab - 4, ab - 2, ab + 1 and ab + 3 have the same tens digit
Two cases
a) The smallest 3 have same ten’s digit => ab-4, ab-2 and ab+1
Let us take smallest possible values for ab. ab+1<20, that is ab<19, but ab+3>19, that is ab>16.
Value of ab could be 17 or 18. So Tens digit could be anything but units digit will be 7 or 8. But the units digit of squares (\(0^2,1^2,2^2,3^2,4^2,5^2,6^2,7^2,8^2,9^2\)can be only 0,1,4,9,6,5,6,9,4,1.
So 7 or 8 not possible.
b) The largest 3 have same ten’s digit => ab+3, ab-2 and ab+1
Let us take possible values for ab...... ab-2>19, that is ab>21, but ab-4<20, that is ab<24.
Value of ab could be 22 or 23. Again 2 or 3 cannot be units digit of a perfect square.
Hence ab is NOT a square, and a is not equal to b.
(2) ab is divisible by 8, but not divisible by 16.
So ab=2^3x, but not 2^4x.
If you have power of 3, it cannot be a square.
D
19 Jan 2021, 08:51
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If a and b are positive integers and ab > 13, is a ≠ b?
Stat1: Exactly three of the numbers ab - 4, ab - 2, ab + 1 and ab + 3 have the same tens digit
If tens digit is same of exactly 3 cases, then last digit will be 3 or 7. if a=b, then last digit won't be 3 or 7. So, Sufficient
Stat2: ab is divisible by 8, but not divisible by 16.
So, last 3 digits will be divisible by 8. we can have ab= 24 = 8*3, 40= 8*5...So, ab=8*odd number, it means "a" can't equal to "b". Sufficient
So, I think D.
Stat1: Exactly three of the numbers ab - 4, ab - 2, ab + 1 and ab + 3 have the same tens digit
If tens digit is same of exactly 3 cases, then last digit will be 3 or 7. if a=b, then last digit won't be 3 or 7. So, Sufficient
Stat2: ab is divisible by 8, but not divisible by 16.
So, last 3 digits will be divisible by 8. we can have ab= 24 = 8*3, 40= 8*5...So, ab=8*odd number, it means "a" can't equal to "b". Sufficient
So, I think D.
