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\(
\text { What is the units digit of } 57^{45} \text { ? }
\)
\(
\text { So, the units digit of } 57^{45} \text { is } 7
\) because the power 45 is in the form of \(4x +1\). According to the table, it should be 7.
\(
\text { Find the unit digit of } 32^{24} \text {. }
\)
\(
\text { Therefore, the unit digit of } 32^{24} \text { is } 6 \text {. }
\) because the power is expressed in the form of 4 x 6. According to the table, the answer is 6.
\(
\text { Find the unit digit of } 32^{27} \text {. }
\)
When 27 is divided by 4, the remainder is 3.
Since it is in the form of 4 x 6 +3, the unit digit should be 8.
Is the number \(595,378,263,068,723,132\) a perfect square?
The unit digit is a 2. Are there any digits which, when squared, end in a 2? There is no number \(n\) for which \(n^2\) ends in a 2, so the number above cannot possibly be a perfect square.
\(
\text { Is } 9,089^2+765^2 \text { equal to } 83,195,146 \text { or } 83,195,140 \text { ? }
\)
\(9089^2\) ends in a 1 and \(765^2\) ends in a 5, so their sum ends in a 1+5=6. Given the choices, 83,195,146 is the answer that makes sense.
\(
\text { What is the units digit of }\left(346^2+364^2\right)^2 \text { ? }
\)
Looking at the units digits \((346^2+364^2)^2=(…..6 + …..6)^2=(……2)^2=….4\)
\(
\text { Given that } 4,624 \text { is a perfect square, what is } \sqrt{4,624} \text { ? }
\)
The unit digit of \(\sqrt{4,624}\) must be 2 or an 8. 62 is too small, so we assume \(\sqrt{4,624}\) must be 68.
\(
\text { What is the units digit of }(1+2+3+4+\ldots+29+30)^2 \text { ? }
\)
(1+2+…..+30) = (30 x 31)/2=15 x 31, which has a units digit of 5. The unit digit remains 5 when squared.
\(
\text { What is the units digit of } 2222^{333} \times 3333^{222} ?
\)
The problem statement asks to find out the units digit of \(2222^{333} \times 3333^{222}\).
Here we can say that the units digit of \(2222^{333}\) is similar to that of \(2^{333}\);
Further, we can say that the units digit of \(3333^{222}\) is similar to that of \(3^{222}\);
Hence, it can be inferred that the units digit of \(2222^{333} \times 3333^{222}\) is the same as that of \(2^{333} \times 3^{222}\);
The units digit of both the numbers 2 and 3 in positive integer power gets repeated in 4 patterns.
In the case of 2 , it is \(\{2,4,8,6\}\) and in the case of 3 , it is \(\{3,9,7,1\}\).
Since the unit digit of \(2^{333}\) will be similar to that of \(2^{1}\), so the unit digit is 2 . This is because if 333 is divided by a cyclicity of 4 , it gives a remainder of 1 . This denotes that the unit digit of \(2^{333}\) is the first number from the pattern.
Since the unit digit of \(3^{222}\) will be similar to that of \(3^{2}\), so the unit digit is 9 .
This is because if 222 is divided by a cyclicity of 4 , it gives a remainder of 2 . This denotes that the units digit \(3^{222}\) is the second number from the pattern.
Therefore, we get \(2 \times 9=18\)
Thus, the units digit of \(2222^{333} \times 3333^{2222}\) is 8 .
Peter multiplies four consecutive integers and divides the result by 10, leaving a remainder. What is the remainder?
In order for there to be a reminder, the units digits of the consecutive integers can only be 1,2,3, and 4 or 6,7,8, and 9. In this case, the unit digits of the product of the integers will be 4. This means that when the integer is divided by 10 the remainder is 4.
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