Patternless?

You are asked for the remainder when divided by 10, which means you are looking for the ones digit of that expression. If that monstracity reduces to 27, then 7 would be the answer; if it reduced to 6,518,309, then 9 would be the answer. Because of this, we're only interested in the ones digit of that obscenely large number.

Now 1 to any power is just 1.

Powers of 8 follow this pattern for their ones digit: 8; 4; 2; 6; 8;. . . . A sequence with 4 repeating terms. 1234657890 leaves 2 left over when divided by 4, so the ones digit of 8^1234567890 is the 2nd term of that repeating sequence, 4

Powers of 3 follow this pattern for their ones digit: 3; 9; 7; 1; 3;. . . . Also a sequence with 4 repeating terms, like above. So, the ones digit of 3^1234567890 is 9, the second term of the the repeating sequence.

Powers of 6 always end in a 6, so the ones digit of 6^1234567890 is 6.

Putting this into the expression, you get (1+~4-~9-~6)^1234567890, where you don't really care what the ~ means, because we're only interested in the ones digit. This simplifies to (~0)^1234567890, which will be 0. So, the ones digit of the expression is 0

${ 1 }^{ n }\quad =\quad 1\quad mod\quad 10$

${ 6 }^{ n }\quad =\quad 6\quad mod\quad 10$

Let

${ 8 }^{ n }\quad =\quad a\quad mod\quad 10$

${ 3 }^{ n }\quad =\quad b\quad mod\quad 10$

Hence, ${ 8 }^{ n }\quad - { 3 }^{ n }\quad$ = a -b mod 10

= 5 mod 10

Also, ${ 1 }^{ n }\quad - { 6 }^{ n }\quad = 1-6 mod 10$

= -5 mod 10

Finally,

${ 1 }^{ n }\quad - { 6 }^{ n }\quad + { 8 }^{ n }\quad - { 3 }^{ n }\quad = 5 -5 mod 10$

= 0 mod 10

hence for every n, the remainder is $\boxed { 0 }$

Well, why not just replace that monstrosity with a smaller number like 2, 3... and so forth? A pattern can be devised as the total will always end with 0.

I assumed 1234567890 as x.Then I keep substituting x=1,2,3,4,5.Every time it was giving 0 as reminder.Hence I concluded '0' by principle of mathematical induction because the answer has to be consistent

You are asked for the remainder when divided by 10, which means you are looking for the ones digit of that expression. If that monstracity reduces to 27, then 7 would be the answer; if it reduced to 6,518,309, then 9 would be the answer. Because of this, we're only interested in the ones digit of that obscenely large number. Now 1 to any power is just 1. Powers of 8 follow this pattern for their ones digit: 8; 4; 2; 6; 8;. . . . A sequence with 4 repeating terms. 1234657890 leaves 2 left over when divided by 4, so the ones digit of 8^1234567890 is the 2nd term of that repeating sequence, 4 Powers of 3 follow this pattern for their ones digit: 3; 9; 7; 1; 3;. . . . Also a sequence with 4 repeating terms, like above. So, the ones digit of 3^1234567890 is 9, the second term of the the repeating sequence. Powers of 6 always end in a 6, so the ones digit of 6^1234567890 is 6. Putting this into the expression, you get (1+~4-~9-~6)^1234567890, where you don't really care what the ~ means, because we're only interested in the ones digit. This simplifies to (~0)^1234567890, which will be 0. So, the ones digit of the expression is 0

1 + 8 - 3 - 6 = 0, power of 0 = 0, 0 / 10 = 0,
So zero is the answer, simple.

If n is a natural number (positive integer) 1^n = 1, 8^n mod 10 = 6 for all n > 2, 3^n mod 10 = 1 for all n > 2, 6^n mod 10 = 6 for all n

So when you have n> 2 and you add (1^n + 8^n - 3^n - 6^n) the last digit will always be 0 and therefore a multiple of 10. The remainder when dividing by 10 will be 0.

hint use 9can be written as -1 similarly use others they will nullify=0

1 + 8 - 3 -6 = 0

So 0 ^ 1234567890 = 0

And 0 / 10 = 0

EASY

Kindly use latex!!! This is not a solution