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The unit digit of the cube of the number 1787 is â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.

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According to the question, we need to find the unit digit in the cube of the number 1787 . Now the digit in the units place in 1787 is 7. Let us first place the digits of 1787 as per the decimal system:

Now, we have 7 in the unit's place in 1787.

So the cube of 1787 means multiplying 1787 with itself thrice, that is,${\left( {1787} \right)^3} = \left( {1787} \right) \times \left( {1787} \right) \times \left( {1787} \right)$ .

So, clearly the digit in the units place being 7 is also being cubed, which means that the digit in the units place in the cube of 1787 will be the same as the digit in the units place in the cube of 7.

So, ${7^3} = 7 \times 7 \times 7 = 343$ .

Thus, the digit in the units place of ${7^3}$ is 3

$ \Rightarrow $ the digit at the units place of ${\left( {1787} \right)^3}$ will be 3.

Hence, the unit digit of the cube of the number 1787 is â€¦..3â€¦..

So, ${\left( {43} \right)^1}$ has $3 \times 1 = 3$ as its unit digit.

${\left( {43} \right)^2}$ has $3 \times 3 = 9$ as its unit digit.

${\left( {43} \right)^3}$ has $3 \times 3 \times 3 = 27$,that is 7 as its unit digit.

${\left( {43} \right)^4}$ has $3 \times 3 \times 3 \times 3 = 81$, that is 1 as its unit digit.

${\left( {43} \right)^5}$, has $3 \times 3 \times 3 \times 3 \times 3 = 243$, that is 3 as its unit digit.

So, as soon as the power increases by 1 from 4 and becomes 5, the unit digit also starts repeating itself from 3 onwards.

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