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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
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A happy number is a positive integer defined in the following way: in
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16 Apr 2021, 02:55
16 Apr 2021, 02:55
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A happy number is a positive integer defined in the following way: in sequence A for integer i > 0, the term \(A_{(i+1)}\) equals the sum of the squares of digits of \(A_i\). If \(A_n = 1\) for some positive integer n, then \(A_0\) is happy. Which of the following is NOT happy?
A. 7
B. 10
C. 13
D. 16
E. 19
A. 7
B. 10
C. 13
D. 16
E. 19
A happy number is a positive integer defined in the following way: in
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03 May 2021, 15:03
03 May 2021, 15:03
Expert Reply
Official Explanation
Since there can only be one right answer, only one of the choices can be unhappy; the other four numbers must be happy. So, as you check those numbers against the definition of happiness, make sure you see the same outcome for four of the numbers.
Start with answer choice (A), which is 7. If this is a happy number, then it should start a sequence that produces 1 eventually. We have the definition of the sequence; let’s apply it.
\(A_0 = 7\)
\(A_1\) = sum of squares of digits of \(A_0 = 7^2 = 49\)
\(A_2\) = sum of squares of digits of \(A_1 = 4^2 + 9^2 = 16 + 81 = 97\)
\(A_3\) = sum of squares of digits of \(A_2 = 9^2 + 7^2 = 81 + 49 = 130\)
\(A_4\) = sum of squares of digits of \(A_3 = 1^2 + 3^2 + 0^2 = 1 + 9 + 0 = 10\)
\(A_5\) = sum of squares of digits of \(A_4 = 1^2 + 0^2 = 1 + 0 = 1\)
You’ve gotten a 1, so now you know that 7 (the number you started with) is happy—and therefore not the right answer.
You can eliminate 10 (choice B) easily, either by directly applying the sequence definition or by observing that you got 10 along the way from 7 to 1, so 10 is happy as well.
Try 13 (choice C):
\(A_0 = 13\)
\(A_1\) = sum of squares of digits of \(A_0 = 1^2 + 3^2 = 1 + 9 = 10\)
You can stop with this sequence now, since you know that 10 is happy. 13 is happy as well.
Try 16 (choice D):
\(A_0= 16\)
\(A_1\) = sum of squares of digits of \(A_0 = 1^2 + 6^2 = 1 + 36 = 37\)
\(A_2\) = sum of squares of digits of \(A_1 = 3^2 + 7^2 = 9 + 49 = 58\)
\(A_3\) = sum of squares of digits of \(A_2 = 5^2 + 8^2 = 25 + 64 = 89\)
\(A_4\) = sum of squares of digits of \(A_3 = 8^2 + 9^2 = 64 + 81 = 145\)
\(A_5\) = sum of squares of digits of \(A_4 = 1^2 + 4^2 + 5^2 = 1 + 16 + 25 = 42\)
\(A_6\) = sum of squares of digits of \(A_5 = 4^2 + 2^2 = 16 + 4 = 20\)
\(A_7\) = sum of squares of digits of \(A_6 = 2^2 + 0^2 = 4\)
\(A_8\) = sum of squares of digits of \(A_7 = 4^2 = 16\)
Uh-oh—you’re back to 16, so the cycle will repeat from here infinitely, never reaching 1. This means that 16 is unhappy, and thus is the answer.
If you check 19, you’ll find that it’s happy (19 → 82 → 68 → 100 → 1).
By the way, this definition of “happiness” really exists in number theory. We do make up some definitions for our problems, but this one’s “real,” so to speak!
The correct answer is D.
Since there can only be one right answer, only one of the choices can be unhappy; the other four numbers must be happy. So, as you check those numbers against the definition of happiness, make sure you see the same outcome for four of the numbers.
Start with answer choice (A), which is 7. If this is a happy number, then it should start a sequence that produces 1 eventually. We have the definition of the sequence; let’s apply it.
\(A_0 = 7\)
\(A_1\) = sum of squares of digits of \(A_0 = 7^2 = 49\)
\(A_2\) = sum of squares of digits of \(A_1 = 4^2 + 9^2 = 16 + 81 = 97\)
\(A_3\) = sum of squares of digits of \(A_2 = 9^2 + 7^2 = 81 + 49 = 130\)
\(A_4\) = sum of squares of digits of \(A_3 = 1^2 + 3^2 + 0^2 = 1 + 9 + 0 = 10\)
\(A_5\) = sum of squares of digits of \(A_4 = 1^2 + 0^2 = 1 + 0 = 1\)
You’ve gotten a 1, so now you know that 7 (the number you started with) is happy—and therefore not the right answer.
You can eliminate 10 (choice B) easily, either by directly applying the sequence definition or by observing that you got 10 along the way from 7 to 1, so 10 is happy as well.
Try 13 (choice C):
\(A_0 = 13\)
\(A_1\) = sum of squares of digits of \(A_0 = 1^2 + 3^2 = 1 + 9 = 10\)
You can stop with this sequence now, since you know that 10 is happy. 13 is happy as well.
Try 16 (choice D):
\(A_0= 16\)
\(A_1\) = sum of squares of digits of \(A_0 = 1^2 + 6^2 = 1 + 36 = 37\)
\(A_2\) = sum of squares of digits of \(A_1 = 3^2 + 7^2 = 9 + 49 = 58\)
\(A_3\) = sum of squares of digits of \(A_2 = 5^2 + 8^2 = 25 + 64 = 89\)
\(A_4\) = sum of squares of digits of \(A_3 = 8^2 + 9^2 = 64 + 81 = 145\)
\(A_5\) = sum of squares of digits of \(A_4 = 1^2 + 4^2 + 5^2 = 1 + 16 + 25 = 42\)
\(A_6\) = sum of squares of digits of \(A_5 = 4^2 + 2^2 = 16 + 4 = 20\)
\(A_7\) = sum of squares of digits of \(A_6 = 2^2 + 0^2 = 4\)
\(A_8\) = sum of squares of digits of \(A_7 = 4^2 = 16\)
Uh-oh—you’re back to 16, so the cycle will repeat from here infinitely, never reaching 1. This means that 16 is unhappy, and thus is the answer.
If you check 19, you’ll find that it’s happy (19 → 82 → 68 → 100 → 1).
By the way, this definition of “happiness” really exists in number theory. We do make up some definitions for our problems, but this one’s “real,” so to speak!
The correct answer is D.
gmatclubot
A happy number is a positive integer defined in the following way: in [#permalink]
03 May 2021, 15:03
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