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What is the digit on the units place in the expanded value of 97^275 –

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Re: What is the digit on the units place in the expanded value of 97^275 –  [#permalink]

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New post 02 Jul 2019, 09:10
Q: Unit digit value for the operation (97)^(275)-(32)^(44)?

Since only the unit digit has been asked - all we have to do is find the unit digit of each (to the power) operation and subtract both the numbers to find the overall unit digit value of the equation.

Things to Do

- Find the unit digit of 97^275
- Find the unit digit of 32^44

1. To find the unit digit of 97^275, it is enough to calculate the last digit of the operation 7^275.

Write down the unit digits of the 7-square tables till you find a repeat pattern. We are interested in finding the last digit only, so i am ignoring the other digits,
7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1
- - - - - -
7^5 = 7
7^6 = 9

From above, we can see that the pattern starts repeating from 7^5 onwards, i.e, the 5th power.
Now that we have a repeating pattern, we find the last digit of 7^275 by finding out how many times does 275 divides 4. Doing the math, this operation gives a remainder 3 (275/4 = 68+3)
So, the last digit of 97^275 is same as 7^3 which is 3

2. To find the unit digit of 32^44, let's follow the same procedure:

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 6
- - - - - -
2^5 = 2
2^6 = 4

Doing 44/4 gives no remainder. Therefore the last digit of 32^44 is 6

Therefore, 6-3 =3.

Answer is B
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What is the digit on the units place in the expanded value of 97^275 –  [#permalink]

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New post Updated on: 02 Jul 2019, 09:21
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The questions tests the knowledge of cyclisity.
Here we are given 2 number but our main focus should be on the last digits of both the numbers, that is, 7 and 2.
Cyclisity of 7 is 4 (7, 9, 3, 1), so in \(7^{275}\), when we divide 275 by 4 and get 3 as a remainder. So the number \(7^{275}\) will end with the last digit of \(7^{3}\) and that is 3.

On the similar lines, the Cyclisity of 2 is 4 (2, 4, 8, 6), so in \(2^{44}\), when we divide 44 by 4 and get 0 as a remainder. When the remainder = 0, the number ends in \(2^{cyclisity}\) = \(2^{44}\), that is 6.

Now we have (Number ending with 3) - (Number ending with 6).
Think of a few examples : 13 - 6 = 7 / 23-16 = 7 / 143-56 = 87 (Units digit is 7).

So, the last digit would be 7. Answer is D.

Relevant Theory -


Quote:
LAST DIGIT OF A POWER

Determining the last digit of \((xyz)^n\):

1. Last digit of \((xyz)^n\) is the same as that of \(z^n\);
2. Determine the cyclicity number \(c\) of \(z\);
3. Find the remainder \(r\) when \(n\) divided by the cyclisity;
4. When \(r>0\), then last digit of \((xyz)^n\) is the same as that of \(z^r\) and when \(r=0\), then last digit of \((xyz)^n\) is the same as that of \(z^c\), where \(c\) is the cyclisity number.

• Integer ending with 0, 1, 5 or 6, in the integer power k>0, has the same last digit as the base.
• Integers ending with 2, 3, 7 and 8 have a cyclicity of 4.
• Integers ending with 4 (eg. \((xy4)^n\)) have a cyclisity of 2. When n is odd \((xy4)^n\) will end with 4 and when n is even \((xy4)^n\) will end with 6.
• Integers ending with 9 (eg. \((xy9)^n\)) have a cyclisity of 2. When n is odd \((xy9)^n\) will end with 9 and when n is even \((xy9)^n\) will end with 1.

Originally posted by aalekhoza on 02 Jul 2019, 09:10.
Last edited by aalekhoza on 02 Jul 2019, 09:21, edited 1 time in total.
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What is the digit on the units place in the expanded value of 97^275 –  [#permalink]

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New post 02 Jul 2019, 09:14
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97^275 - 32^44

7 to the power runs in cycle of 7,9,3,1 unit digit- in cycle of 4 numbers
2 to the power runs in cycle of 2,4,8,6 unit digit

So when divided by 4 with the remainder we can tell the units digit
so 275/4 => 3 remainder , so its unit digit is third term in the cycle - 3
44 is divisible by 4 ,so it finishes the cycle==> 6 is units digit

==> 3-4 = 7 will be the units digit of the given number - Option D
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What is the digit on the units place in the expanded value of 97^275 –  [#permalink]

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New post 02 Jul 2019, 09:15
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7 has the cycle of repetition after ^4 and similarly, 2 has the cycle of repetition after^4.

So after resolving it comes out to be ..7^3 - ..2^4 ------> 3-6 ------> 7

So the option is D
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Re: What is the digit on the units place in the expanded value of 97^275 –  [#permalink]

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New post 02 Jul 2019, 09:20
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The units digit of the powers of 7 follow the pattern of 7, 9, 3, 1, 7, 9 and goes on. To note that, for every 4th power the cyclic order remains the same. Therefore, when 275 is divided by 4, we get remainder as 3. The corresponding units digit will be 3. (The Tenths digit can be any number)
Similarly, the unit digits of the powers of 2 follow the pattern of 2, 4, 8, 6, 2, 4 and so on. Deducting the same way, we find that the units digit will be 6.
Since 6 cannot be subtracted from 3, the final answer has to be 13 - 6 = 7 (3 borrows 1 from tenths digit)
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Re: What is the digit on the units place in the expanded value of 97^275 –  [#permalink]

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New post 02 Jul 2019, 09:20
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Solution :
The units digit of \(97^{275}\)can be determined by determining \(7^{275}\), Here we need to check the pattern of 7. I.e \(7^1\) is 7 , \(7^2\) is 49 ,\(7^3\) is 343 & \(7^4\) is 2401 so the pattern of units digit is 7,9,3, & 1. However, when we divide the number 275 by 4, we get the remainder as 3 hence the units digit should be the 3rd number of the pattern hence 3,
Likewise, if we calculate for \(32^{44}\) , we need to identify the pattern of 2, which is 2,4,8,6. 44 however is divisible by 4 so here the units digit must be 6. if we do the subtraction here, unit digit of the first number that is 3 minus that of the second number, the answer comes a 7 in the units place Hence D is the correct answer
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Re: What is the digit on the units place in the expanded value of 97^275 –  [#permalink]

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New post 02 Jul 2019, 09:22
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Units digit of 7^1 - 7.
Units digit of 7^2 - 9
Units digit of 7^3 - 3
Units digit of 7^4 - 1

=> 97 ^ 275. 97^(4n+3) => Unit digit ends with 3

Units digit of 2^1 - 2
Units digit of 2^2 - 4
Units digit of 2^3 - 8
Units digit of 2^4 - 6

=> 32 ^ 44. 97^(4n) => Unit digit ends with 6.

3 - 6 => 13 - 6 = 7

Ans D
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Re: What is the digit on the units place in the expanded value of 97^275 –  [#permalink]

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New post 02 Jul 2019, 09:31
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97 to the power 275. Last digit has cyclicity of 4 ie. 7,9,3,1. 275 is 4n+3. Hence last digit is 3

Second term also has cyclicity of 4. 2,4,8,6. 44 is divisible by 4. Hence last digit is 6

So it's basically 3 and 6 as last digits.
13-6=7

Answer is 7

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Re: What is the digit on the units place in the expanded value of 97^275 –  [#permalink]

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New post 02 Jul 2019, 09:32
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97 to the power 275. Last digit has cyclicity of 4 ie. 7,9,3,1. 275 is 4n+3. Hence last digit is 3

Second term also has cyclicity of 4. 2,4,8,6. 44 is divisible by 4. Hence last digit is 6

So it's basically 3 and 6 as last digits.
13-6=7

Answer is 7

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Re: What is the digit on the units place in the expanded value of 97^275 –  [#permalink]

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New post 02 Jul 2019, 09:34
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The last digit of \(97^275\) is \(3\).

\(7^1 = 7\)
\(7^2 = 9\)
\(7^3 = 3\)
\(7^4 = 1\)

Since the cycle equals \(4\), we need \(275/4\). The remainder is \(3\). Hence \(7^3 = 3\)

The last dgit of \(32^44\) is \(6\)

\(2^1 = 2\)
\(2^2 = 4\)
\(2^3 = 8\)
\(2^4 = 6\)

Since the cycle equals \(4\), we need \(44/4\). There is no remainder. Hence \(2^4 = 6\)

hence,\(...3 - ...6 = ...7\)

Thus D
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What is the digit on the units place in the expanded value of 97^275 –  [#permalink]

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New post 02 Jul 2019, 09:34
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\(97^(275)–32^(44)\) unit digit.

unit digit :\(7^1=7\) , \(2^1=2\)
\(7^2=9\) , \(2^2=4\)
\(7^3=3\),\(2^3=8\)
\(7^4=1\) , \(2^4=6\)
When divided by 275 has remainder 3
so, \(97^(275) =XXXXXXXX3\)
\(32^(44)=YYYYYYY6\)
subtracting them unit digit will be 7
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Re: What is the digit on the units place in the expanded value of 97^275 –  [#permalink]

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New post 02 Jul 2019, 09:38
power of 7 is cycle and power of 2 is same cycle

3-2 is the last two numbers of each powers and results 1
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What is the digit on the units place in the expanded value of 97^275 –  [#permalink]

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New post 02 Jul 2019, 09:39
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\([m]97^{275}\)–\(32^{44}\)[/m]
7 has a round multiplicity of 4
hence divide 275 by 4.. leaving a remainder of 3=
thus unit power of\(97^{275}\) = unit value of\(7^3\)= 3
2 again has a round unit multiplicity of 4
44 will have same as 44/4 =4
\(2^4\) =6
13-6 = 7 . We can safely assume 1 borrowed from next digit
Hence D
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Re: What is the digit on the units place in the expanded value of 97^275 –  [#permalink]

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New post 02 Jul 2019, 09:52
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Point to remember:--The cyclicity of every number from 1 to 9 is 4.

Take 97^275 , unit digit will be the unit digit of 7^275 . Now since the cyclicity is 4 , the unit digit will be of 7^3(275/4).
Therefore the unit digit is 3

Similarly unit digit of 32^44 will be the unit digit of 2^44. The unit digit is thus 6

Difference 3-6 =-3 . But unit digit cannot be negative so we add 10 always when we get such negative values.

Therefore unit digit is -3+10=7

IMO D
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What is the digit on the units place in the expanded value of 97^275 –  [#permalink]

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New post Updated on: 04 Jul 2019, 21:46
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Cyclicity of 7 : 7, 9 , 3, 1 (four)
Cyclicity of 2 : 2, 4 ,8, 6 (four)

275/4 => Reminder : 3
So, last digit of 97^275 = 3 (as per the cyclicity)

Similarly,
44/4 => Reminder : 0
So, last digit of 32^44 = 6

(XXXX3)-(xxxx6) = So, last digit of the result will be 7

Option D seems correct.

Originally posted by sj24 on 02 Jul 2019, 09:54.
Last edited by sj24 on 04 Jul 2019, 21:46, edited 1 time in total.
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Re: What is the digit on the units place in the expanded value of 97^275 –  [#permalink]

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New post 02 Jul 2019, 10:00
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Apply the power 4 rule here, 97^275 its unit digit will be equal to 7^275, divide 275 by 4, we get remainder 3 now, 7^3 the unit digit is 3.
same for 32^44, but here the remainder is 0 when 44 is divided by 4, so replace the power by 4 i.e 2^4 unit digit 6. Now 3-6, take a borrow from the tens digit of 97^275(we need not to know what is it) it will be 13-6 = 7.
Hence D
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Re: What is the digit on the units place in the expanded value of 97^275 –  [#permalink]

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Re: What is the digit on the units place in the expanded value of 97^275 –  [#permalink]

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New post 02 Jul 2019, 10:07
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7^(275) will give 3 in last digit :7 follows a last digit cycle of 4(7,9,3,1) so divide 275/4 R=3 (3 term of cycle )
2^(44) will give 6 as last digit : 2 follows a last digit cycle of 4(2,4,8,6) so divide 44/4 R=0 (4th term of the cycle )

subtracting last digits x3-x6 : 7 at end
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Re: What is the digit on the units place in the expanded value of 97^275 –  [#permalink]

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New post 02 Jul 2019, 10:08
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The answer is D. This is the explanation:
Unit digit for 97^275 can be obtained by multiplying the 7 for 275 times using only the unit digit:
7^1 = 7
7^2 = 49 unit is 9
7^3 = 63 unit is 3
7^4 = 3 * 7 = 21 unit is 1
7^5 = 1 * 7 = 7
7^6 = 7 * 7 = 49 unit is 9
7^7 = 9 * 7 = 63 unit is 3
7^8 = 3 * 7 = 21 unit is 1
Meaning that at every multiple of 4 the unit digit is 1
Thus, 272 is multiple of 4
Therefore; 97^272 = unit digit of 1
97^273 = unit digit of 7
97^274 = unit digit of 9
97^275 = unit digit of 3
Thus 97^275 has unit digit of 3

Similarly, 32^44 will go through the same process using 2
2^1 = 2
2^2 = 2 * 2 = 4 is the unit digit
2^3 = 4 * 2 = 8 is the unit digit
2^4 = 8 * 2 = 16, 6 is the unit digit
2^5 = 6 * 2 = 12, 2 is the unit digit
2^6 = 2 * 2 = 4
2^7 = 4 * 2 = 8
2^8 = 8 * 2 = 16, 6 is the unit digit
This means that at every multiple of 4, 6 is the unit digit.
44 is a multiple of 4 meaning that 32^44 has unit digit of 6

Furthermore; 97^275 - 32^44 = 3 - 6 (using only the unit digits)
Thus, borrow 1 to add to 3 it becomes 13 - 6 = 7
Final answer = 7, ans = D.

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Re: What is the digit on the units place in the expanded value of 97^275 –  [#permalink]

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New post 02 Jul 2019, 10:17
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97^275–32^44

Cyclicity of 97's unit digit 7 is 4 (7,9,3,1)
Cyclicity of 32's unit digit 2 is 4 (2,4,8,6)

For 7^275 unit digit will be 3 [275 leaves remainder 3 when divided by 4]

For 2^44 unit digit will be 6 [44 is divisible by 4].

We need to subtract 6 from 13 (as 3<6) Hence 7

Ans D

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Re: What is the digit on the units place in the expanded value of 97^275 –   [#permalink] 02 Jul 2019, 10:17

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