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

1⁴+6⁴+k⁴+4⁴=1604+10k
1+1296+k⁴+256=1604+10k
k⁴-10k=1604-1553=51
By replacing the options with k, 3 would be the answer.
3⁴-30=51
Option B

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digit 16k4 ; 1^4+6^4+k^4+4^4
so test with each option value
at k=3 we get 1634 =1^4+6^4+3^4+4^4
IMO B

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 ?


this question can be solved just under 30 seconds or may eat up 2 mins as the timer suggests

According to the statement : any AMis given by an n-digit number that is equal to the sum of the nth powers of its individual digits
so for 16k4

1^4 +6^4 +K^4+ 4^4=16k4 : recognize the question is asking for sum of units digit to be 4
so find units digit of each expression and then add up to 4
units digit of 1^4 =1
units digit of 6^4 = 6
units digit of 4^ 4 = 6
so if we add unit digit of each expression we get 13
therefore 3 at units digit + units digit of K^4 = 4

therefore units digit of k^4 has to be 1
now just look through choices only 3 gives a units digit of 1
A. 2
B. 3
C. 4
D. 5
E. 6


though the explanation is big but if you just understand what i did its just a matter of 40 seconds max to get through this daunting question and save time

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1,6k4

\(1^4\)+\(6^4\)+\(k^4\)+\(4^4\) = 1,6k4

\(1^4\)+\(6^4\)+\(4^4\) = 1553

Since we know that we are looking for a number in 1600's. So we need an option when raised to the power of 4 should be greater than 47 and less than 147. Because less than 47 will be in 1500's and greater than or equal to 147 would be in 1700's.

\(2^4\) is 16 and \(4^4\) is 256. So only 3 when raised to the power of 4 is greater than 47 and less than 147.

Lets check 3 - \(3^4 \)= 81. Adding 81 to 1553 gives us 1634, and 10th digit is 3 which is our answer.

Ans: B

add everything except k and you get 1553
16k4-1553 = k^4

only k = 3 works
1634-1553 = 81
3^4 = 81

Solution:

Since 1,6k4 is a 4-digit number, we can create the equation:

1^4 + 6^4 + k^4 + 4^4 = 1000 + 600 + 10k + 4

1 + 1296 + k^4 + 256 = 1604 + 10k

k^4 = 51 + 10k

We see that k must be 3 since 3^4 = 81 and 51 + 10(3) = 81.

Answer: B

The key here is to focus on the units digits:

1^4 = 1
6^4 = 6
4^6 = 6

6 + 6 + 1 = 13.

In 1,6k4 we see that the units digit is 4. What number to the 4th power will give us a units digit of 1? The answer is 3. 3^4 = units digit of 1.

Answer is B.

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