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GMAT 1: 570 Q43 V26 GMAT 2: 660 Q48 V34 Re: What is the digit on the units place in the expanded value of 97^275 –  [#permalink]

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Cyclic nature of 7 and 2 must be known.

7's cyclicity is -> 7,9,3,1 => 4 , 275%4 = 3 so 7^3 will be 3
And 2^44 will, likewise be 6 ( cyclicity 4) 3-6 will be (13-6 , borrow from tens place) which will be 7.

Regards,
Rishav
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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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Hi,
To find the unit digits, we need to consider only the unit of the base and last two digits of power.
Another rule is power raised to certain number work in a cycle that means after some powers it will repeat the same digits.
Thus,
$${97^{275}}$$ = $${7^{75}}$$
using cycle formula => $${a^{4k+1}}$$ = $${a^1}$$, where a is base and k is some constant:
$${7^{75}}$$ = $${7^{3}}$$ => 3 (as 75 = 4*18 + 3)

For, $${2^{44}}$$ => $${2^4}$$ => 6 (as $${a^{4k}}$$ = $${a^4}$$ and 44 = 4*11)

So, as per question :
3-6 = 7 (13-6)
so unit digit is 7.

Please hit kudos if you like the solution.
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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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This question tests your knowledge of the concept of cyclicity.

Since we're only concerned with the unit's place, it's best to rewrite this as 7^275 - 2^44 (since the tens digit would not have any role to play in determining the units place)

Now, 275 = 4*68 + 3 can be written in the form 4k + 3, and similarly, 44 = 4*11 can be written as 4n

Now, we know that units digit cyclicity of both 7 and 2 is 4, i.e. they repeat their units digit after 4. Knowing this, we get the units digit for 7^(4k+3) to be 3 (eg. 7,49,343) and units digit for 2^(4n) to be 6 (2,4,8,16)

Now, subtracting 6 from 3, we get 7 as the units digit and that's our answer (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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(1) All UNIT DIGITS FOLLOW "THE CYCLICITY OF 4" i.e. after every four powers, UNIT DIGIT REMAINS SAME.
So, Divide the all POWERs by 4 and work with the REMAINDER. (IF the REMAINDER is ZERO, take 4 as the remaining number because you have divided the numbers by 4)

A SHORTCUT regarding the DIVISION OF 4: JUST take LAST TWO DIGIT & DIVIDE them by 4.
For example, 275/4 is same as 75/4 (REMAINDER=3).

(2) When you are asked to find out UNIT DIGIT, work with UNIT DIGIT only (CROSS OUT TENS & HUNDREDS).
For example, UNIT DIGIT of (97^275) and UNIT DIGIT of (9^275) are the SAME.

considering cyclic of 4 & unit digit only, the question {what is the unit digit of( 97^3)-(32^4) becomes what is the unit digit of (7^3)-(2^4)?
Here, UNIT DIGIT of 7^3=3 and
UNIT DIGIT of 2^4=6 [NOTE: After dividing 44 by 4, REMAINDER is ZERO, for UNIT DIGIT CYCLICITY PURPOSE we will take 4 as remaining number because remainder must be an integer between 1&4).

NOW, CHECK whether (97^275) IS GREATER THAN (32^44)?
Case 1: if (97^275) IS GREATER THAN (32^44)?, the value of (97^275)- (32^44)=*****************3-************6= ****************7 (because we consider 3 as 13 because first term is GREATER)
Case 1: if (97^275) IS LESS THAN (32^44)?, the value of (97^275)- (32^44)=*****************3-************6=- **********3 (because simply 6 MINUS 3 because SECOND term is GREATER)

CHECKING:
97>32 &
275>44.
SO, (97^275) IS DEFINITELY GREATER THAN(32^44).
SO, ONLY CASE 1 POSSIBLE.

so, UNIT DIGIT=7 (D is the ANSWER)

Originally posted by BelalHossain046 on 02 Jul 2019, 08:32.
Last edited by BelalHossain046 on 02 Jul 2019, 09:29, 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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What is the digit on the units place in the expanded value of 97^275–32^44?

Since we are only asked about the unit digit of the above expression, no. cyclicity principle will help.

7 has a cyclicity of 4- 3,9,7,1- divide 275 by 4 and we get the remainder 3 so the unit digit of the expression 97^275 will be 3.
Same way, 2 has a cyclicity of 4- 2,4,6,8- divide 44 by 4 and there is no remainder, so the unit digit is 6.

Try some nos. 13-6 will have unit digit 7, 23-6 will have unit digit 7 and so on.
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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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(97^275) - 32^44
We interested in only the last digit, so:
7^275 - 2^44
7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1
7^5 = 7

So this pattern repeat, that's mean:
272 is the last multiple of 4 before 275
Number-----273--274--275
LastDigit---7----9----3
Ok, last digit of 7^275 is 3

Than make the same analyze with 2^44
last digit is 6

So 3 - 6 = -3 this can't becase (97^275) > 32^44, so
13 - 6 = 7

Answ 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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To find the unit digit of the given number we need to know the cyclicity of 7 and 2 and that is 4 numbers after that it repeats the same numbers.
7: 7-9-3-1
2 : 2-4-8-6
Therefore unit digit of 97^275 is 3 and unit digit of 32^44 is 6
3-6 = 7
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Tashin

Originally posted by Tashin Azad on 02 Jul 2019, 08:41.
Last edited by Tashin Azad on 02 Jul 2019, 11:13, 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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97^275 - 32^44. Lets assume this to be A - B

To find the units digit of a A-B, we must first know the unit's digit of A and B respectively.

Unit's Digit of A - 97^275 depends on the unit's digit when obtained from 7^275. Since 7 has a cyclicity of 4(7,9,3,1) and 275 = 4(68) + 3 => the unit's digit of A will be 3

Similarly, Unit's Digit of A - 32^44 depends on the unit's digit when obtained from 2^44. Since 2 has a cyclicity of 4(2,4,8,6) and 44 = 4(11) => the unit's digit of A will be 6

Unit's digit of A-B = 3 - 6 = 13 (By Borrowing from the Ten's digit in A) - 6 = 7

Hence, The answer must be D
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The units digit of 97^275 will be 3 and the units digit of 2^44 will be 6. So the units digit of the difference then 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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97^275:
Use only last digit to get the units place:

Power1 - 7^1 = 7
Power2 - 7x7 = 49
Power3 - 9x7 = 63
Power4 - 3x7= 21
Power5 - 1x7 = 7 .... same as 1

so units digit will cycle through 7, 9, 3, 1... four unique digits repeating

275 divde by 4 gives quotient 272 and remainder 3 - so units digit will be 3 (3rd in the cycle)

32^44:
Same logic as above
Power1 - 2^1 = 2
Power2 - 2x2 = 4
Power3 - 4x2 = 8
Power4 - 8x2 = 16
Power5 - 6x2 = 12 .... same units digit as 1

so the cycle here is 2, 4, 8, 6.... again cycle of 4 digits

44 is divisible by 4 - so 44th power of 2 will have units digit 6 (4th in cycle)

Units digit of difference will be 3 - 6 = 13-6 (Do manual subtraction on paper - 1 will be carried to 3 to make it 13) = 7

Ans is D - 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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What is the digit on the units place in the expanded value of 97^275–32^44?

A. 1
B. 3
C. 5
D. 7
E. 9

This can be done with the help of cyclicity of 7 and 2 which is 4. Question can be rephrased as what is the unit digit of 7^275 - 2^44
on dividing 275 with 4 (cyclicity of 7) we get remainder as 3
and dividing 44 with 4 (cyclicity of 2) we get remainder as 0 which will be equal to 4 itself
Unit digit of 7^3 - 2^4 = 3 - 6 = 7 (at one's place)
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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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Concept tested:
This question is based on cyclicity.
The cyclicity of 7 is 7,9,3,1 and that of 2 is 2,4,8,6.
Soln:
Now, if we only see the units digit and the power. 7 is raised to 275.
275 when divided by 4 has quotient 68 and leaves remainder 3. What is means is 7 completes 68 cycles of 7,9,1,3 and then 3 units are left. So taking the 3rd digit in cyclicity, we have the units digit of 37^275 as 3.
Applying the same concept to 2, we come to the conclusion that it completes 11 cycles and leaves no remainder. Hence units digit is the 4th digit in the cyclicity - 6.
Now just consider the first 2 digit number that ends with 3 and subtract 6 from it. This gives you the answer 7. [D]
Note: we can't subtract 3 from 6 and say the remainder is -3. So we take a number that is greater than 6 and ends with 3.
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What is the digit on the units place in the expanded value of 97^275 –  [#permalink]

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we need to find the unit digit of
97^275 - 32^44

so we will consider unit digits of both
$$7^1$$=7
$$7^2$$=9 (unit digit)
$$7^3$$=3
$$7^4$$=1
$$7^5$$=7
$$7^6$$=9
$$7^7$$=3
$$7^8$$=1
from here we conclude that it follows a cyclic pattern $$7^4$$,$$7^8$$ = each unit digit = 1
we need to find unit digit of 7^275
so we will write this in terms of $$7^4$$ -- (7^4)^68 * $$7^3$$ = 1*$$7^3$$ = 3 (unit digit of 97^275)

now unit digit of 2^44
we need to find cyclic pattern
by performing the same operation above on 2 we find pattern $$2^4$$,$$2^8$$ = each unit digit = 6
so we will write 2^44 in terms of $$2^4$$-- (2^4)^11 = 6^11 = 6 (unit digit of 32^44)

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

(...........3) - (....6) = 7 (since 97^275 is larger value than 32^44)

correct answer is 7 option D

Originally posted by shridhar786 on 02 Jul 2019, 08:55.
Last edited by shridhar786 on 02 Jul 2019, 08:57, edited 1 time in total.
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Here, concept of cyclicity is tested. They gave us these huge, ugly looking numbers to distract and make us panic. But we won't. For 97^275 it is enough to know units digit of 7^275. Cyclicity of 7 is 4. that is
7^1=07
7^2=49
7^3=_43
7^4=__01
So, units digit when 7 is raised to the power of 275 is 43.
Likewise, we need to know remainder of 2^44. 2 has also cyclicity of 4, that is
2^1=02
2^2=04
2^3=08
2^4=16,
So, units digit when 2 is raised to the power of 44 is 6, So 43-6=_7 (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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What is the digit on the units place in the expanded value of 97^275–32^44?

A. 1
B. 3
C. 5
D. 7
E. 9

We can get the series of unit places for all powers of 7 & 2 to see the series of repetition.
7^1 = unit digit 7
7^2 = unit digit 9
7^3 = unit digit 3
7^4 = unit digit 1
7^5 = unit digit 7 , So after 4 it gets repeated. Same with 2, after 4, unit digits get repeated.

Now if we calculate for power 275 for 7, 3 is remainder and the unit digit should be "3", and for power 44 for 2, remainder is 0, and unit digit will be "6".
Now unit digit(3) - unit digit(6) = 13-6 = 7.
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Any power of a number ending with 7 will have the units digit either 7,9,3 or 1 and any number ending with 2 will have the units digit either 2,4,8 or 6. Therefore 97^275 and 32^44 will be ending with 3 and 6, giving the answer as 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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Both 7&2 has power cycle of 4

Units digit of 3-6 =7

Hence option 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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This can be simplified as units digit of 7^275 - 2^44

We know periodicity of 7 is 4 ie. (7^4)^n will return 1 in units digit for all n>=1. Now 275 = 4 x 68 + 3, so the units digit is determined by units digit of 7^3 ie. 3.

Again we know We know periodicity of 2 is 4 ie. (2^4)^n will return 6 in units digit for all n>=1. Now 44 = 4 x 11, so the units digit always 6.

Units digit 3 - Units digit 6 = Units digit 7

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

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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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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 What is the digit on the units place in the expanded value of 97^275 –   [#permalink] 02 Jul 2019, 09:14

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