Hi, could anyone give me a better explanation to this problem?:
If a is a positive and if the unit digit of a^2 is 9 and (a+1)^2 is 4, what is (a+2)^2
Thanks!
don't understand the explanation in OG to this problem
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I assume the question asks for the unit digit of (a+2) ^ 2.
if unit digit of a ^ 2 is 9 then unit place a should be 3 or 7
if unit digit of (a+1) ^ 2 is 4 then unit place can only be 8
=> unit digit of (a+2) ^ 2 is 1
Hope this helps.
if unit digit of a ^ 2 is 9 then unit place a should be 3 or 7
if unit digit of (a+1) ^ 2 is 4 then unit place can only be 8
=> unit digit of (a+2) ^ 2 is 1
Hope this helps.
can you please elaborate... your analysis seems to have a ton of holes... how do the first two statements conclude into the third one!?rajs.kumar wrote:I assume the question asks for the unit digit of (a+2) ^ 2.
if unit digit of a ^ 2 is 9 then unit place a should be 3 or 7
if unit digit of (a+1) ^ 2 is 4 then unit place can only be 8
=> unit digit of (a+2) ^ 2 is 1
Hope this helps.
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Raj is right. If the units digit of a^2 is 9, only two numbers will square to give a units digit of 9: 3 or 7. See below.
Units digit when squaring:
0^2 = 0
1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 6
5^2 = 5
6^2 = 6
7^2 = 9
8^2 = 4
9^2 = 1
So, a is either 3 or 7. If (a+1)^2 units digit is 4, then a must be 7, because (7+1)^2 = 8^2 = units digit of 4. a can't be 3, b/c (3+1)^2 = 4^2 = units digit of 6.
So if a is 7, then (a+2) = 9. 9^2 = units digit of 1.
Units digit when squaring:
0^2 = 0
1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 6
5^2 = 5
6^2 = 6
7^2 = 9
8^2 = 4
9^2 = 1
So, a is either 3 or 7. If (a+1)^2 units digit is 4, then a must be 7, because (7+1)^2 = 8^2 = units digit of 4. a can't be 3, b/c (3+1)^2 = 4^2 = units digit of 6.
So if a is 7, then (a+2) = 9. 9^2 = units digit of 1.
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