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An “Armstrong number” is an n-digit number that is equal to

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by BTGModeratorVI » Sun Jul 26, 2020 6:42 am

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An “Armstrong number” is an n-digit number that is equal to the sum of the nth powers of its individual digits. For example, 153 is an Armstrong number because it has 3 digits and 1^3 + 5^3 + 3^3 = 153. What is the digit k in the Armstrong number 1,6k4 ?

A. 2
B. 3
C. 4
D. 5
E. 6
Answer: B
Source: Official guide
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Re: An “Armstrong number” is an n-digit number that is equal to

by Jay@ManhattanReview » Sun Jul 26, 2020 10:47 pm
BTGModeratorVI wrote: ↑
Sun Jul 26, 2020 6:42 am
 An “Armstrong number” is an n-digit number that is equal to the sum of the nth powers of its individual digits. For example, 153 is an Armstrong number because it has 3 digits and 1^3 + 5^3 + 3^3 = 153. What is the digit k in the Armstrong number 1,6k4 ?

A. 2
B. 3
C. 4
D. 5
E. 6
Answer: B
Source: Official guide
So, as per the given information, we have 1,6k4 = 1^4 + 6^4 + k^4 + 4^4

1,6k4 = 1 + 1,296 + k^4 + 256

1,6k4 = 1,553 + k^4

1,000 + 600 + 10k + 4 = 1,553 + k^4

1,604 – 1,553 + 10k = k^4

51 + 10k = k^4

Now solving this higher-order equation is a tedious task. Let's make things easier by analyzing options.

Note that irrespective of the value of k, the units digits of 10k would be 0. Thus, the units digits of 51 = 10k would be 1. This means that the units digits of k^4 is also 1.

Only option B (= 3) qualifies as 3^4 = 81, the units digits = 1.

Correct answer: B

Hope this helps!

-Jay
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Re: An “Armstrong number” is an n-digit number that is equal to

by swerve » Mon Jul 27, 2020 4:31 am
BTGModeratorVI wrote: ↑
Sun Jul 26, 2020 6:42 am
 An “Armstrong number” is an n-digit number that is equal to the sum of the nth powers of its individual digits. For example, 153 is an Armstrong number because it has 3 digits and 1^3 + 5^3 + 3^3 = 153. What is the digit k in the Armstrong number 1,6k4 ?

A. 2
B. 3
C. 4
D. 5
E. 6
Answer: B
Source: Official guide
\(1^4+6^4+k^4+4^4=1604+10k\)
\(1+1296+k^4+256=1604+10k\)
\(k^4-10k=1604-1553=51\)

By replacing the options with \(k\), 3 would be the answer.

\(3^4-30=51\quad \Longrightarrow\quad\) B
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Re: An “Armstrong number” is an n-digit number that is equal to

by Brent@GMATPrepNow » Mon Jul 27, 2020 5:46 am
BTGModeratorVI wrote: ↑
Sun Jul 26, 2020 6:42 am
 An “Armstrong number” is an n-digit number that is equal to the sum of the nth powers of its individual digits. For example, 153 is an Armstrong number because it has 3 digits and 1^3 + 5^3 + 3^3 = 153. What is the digit k in the Armstrong number 1,6k4 ?

A. 2
B. 3
C. 4
D. 5
E. 6
Answer: B
Source: Official guide
1,6k4 is a 4-digit number.
So, 1⁴ + 6⁴ + k⁴ + 4⁴ = 16k4
Evaluate: 1 + 1296 + k⁴ + 256 = 16k4
Simplify: 1553 + k⁴ = 16k4

Whatever k is, it must be the case that the UNITS digit of k⁴ is 1, so that 1553 + k⁴ = 16k4
Test some values..
k = 1: 1⁴ = 1, so we get: 1553 + 1 = 1554 NO GOOD
k = 3: 3⁴ = 81, so we get: 1553 + 81 = 1634 WORKS!!

Answer: k = 3

Cheers,
Brent Hanneson - Creator of GMATPrepNow.com
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