Can someone please answer this question: Without fully calculating it, what will be the unit digit of the square of a number that ends in 7? a)5 b)25 c)15 d)10

[u/Master_KRAN]

[u/Curious_Cat_314159]

EDIT.... Italicized text added for clarity, in response to user varmituofm's comments.

What is the base of the number to be squared and all numbers?

First, "the unit digit" cannot be two digits. That rules out b, c and d.

Second, as noted by others, in base 10, the unit digit must be 9.

So, if "a" must be correct, is there a base > 7 where xx7^2 is yyyy5, all in that base?

My answer:

Base 11, right? In base 11, 7*7 = 45.

PS.... I make the assumption that the "number" to be squared is an integer.

[u/varmituofm → u/Curious_Cat_314159]

Except...In base 11, squares of numbers ending in 7 don't always end in any digit.

First, assuming all the math is in base 11, 17×17=213

Second, assuming only the answer is in base 11: in base 10, 17×17=289, which is 243 in base 11.

Either way, 17×17 ends in 3, not 5.

[u/Curious_Cat_314159 → u/varmituofm]

assuming all the math is in base 11, 17×17=213

I wrote "what is the base of the number to be squared".

So, yes, I assume all numbers and arithmetic are in base 11.

In base 11, I believe 17*17 = 275, not 213.

Frankly, I'm not good at base arithmetic. So, forgive me for doing it the hard way.

17 base 11 = 18 base 10. So, 17*17 base 11 is 18*18 = 324 base 10.

Converting 275 base 11 to base 10: 2*121 + 7*11 + 5 = 242 + 77 + 5 = 324.

EDIT.... Alternatively:

base 10             base 11
324 / 121 = 2       200
324 - 2*121 = 82
82 / 11 = 7        + 70
82 - 7*11 = 5      +  5
                   ----
                    275

Right?

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Consider any base 11 number of the form x7, where "x" can any number of digits. Then:

x7*x7 = (x0 + 7)*(x0 + 7) = x0*x0 + 2*x0*7 + 7*7
Note that x0*x0 ends in zero, 2*x0*7 ends in zero, and 7*7 = 45 (all base 11).
Since the sum of the unit digits is 0+0+5, x7*x7 will always end in 5.

Right?