Question

1003 one dight answer

Answer 1

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Answer 2

3^1 = 3
3^2 = 9 , 3^3 = 27 , 3^4 = 81 and 3^5 = 243
so, 3^1000 = (3^4)^250 = (81)^250
Now, as the ones digit is 1 of 81. so, 81^250 will have 1 as in its ones place.
Now, 81^250 = 3^1000.
so, 3^1000 will have 1 in its ones place.
Now, 3^1001 = 3^1000 × 3
so, 3^1001 will have 3 as its ones digit.

Now, 7^1 = 7 , 7^2 = 49 , 7^3 = 343 and 7^4 = 2401
Now, 7^1002
= 7^1000×7^2
= (7^4)^250 × 7^2
= (2401)^250×49
Now, (2401)^250 will end with the digit 1.
so, (2401)^250×49 will end with the digit 9.
Now, (2401)^250×49 = 7^1002
so, 7^1002 will end with the digit 9 at its ones place.
(1×49 = 49 ; so, it will end with the digit 9)

Again, 13^1 = 13,13^2 = 169 , 13^3 = 2197 and 13^4 = 28541
Now, 13^1003 = 13^1000×13^3
= (13^4)^250 × 13^3
= (28541)^250 × 2197
Now, (28541)^250 will end with the digit 1 and so, when it will be multiplied by 2197, the resulting digit at ones place be 7
so,(28541)^250 × 2197 will end with the digit 7
=》13^1003 will end with the digit 7.

so, 3^1001 will end with the digit 3 , 7^1002 will end with 9 and 13^1003 will end with the digit 7.
Now, 3^1001 × 7^1002 × 13^1003
will end with the digit which comes at the ones place of the product of 3,9 and 7
Now, (3×9×7) = 189 which has 9 at its ones digit place.
so, 3^1001 × 7^1002 × 13^1003 will end with the digit 9.