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We have to find unit digit of 97^275–32^44
275=3 mod 4
Hence unit digit of 97^275 is same as 7^3=3

44=0 mod 4
Hence unit digit of 32^44 is same as 2^4=6

unit digit of 97^275–32^44= x3-y6=7
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The cyclicity of 7 and 2 is 4.That is the last digit repeats after every four powers.

Say for 7
7
9
3
1
again
7
9
3
1
Similarly for 2 it is 2,4,8,6

Now
97^275 implies last digit is 7 ^ 3= 3 (275/4 leaves 3 )

32^44 implies last digit is 6 (44/2 leaves 0 )

Therefore last digit = 3-6 = 13-6= 7 (last digit will not be negative we will have to borrow one )
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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

In 97^275, unit digit will depend on 7^275 and in 32^44 will depend on 2^44.
Unit digit of
7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1
7^5 = 7
In general unit digit of 7^(4x)=1
275=4*68+3
Unit digit of 97^275 = 3

Similarily,
Unit digit of
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 6
2^5 = 2
In general unit digit of 2^(4x)=6
44=4*11
Unit digit of 32^44 = 6

Unit digit of 97^275-32^44 = 13- 6 = 7 since 3-6 gives negative value and one is borrowed from 10's digit.

IMO D
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Think answer is D.

97^275 - 32^44

7^1=7
7^2=9
7^3=3
7^4=1
7^5=7 .. the trend continues 7,9,3,1

2^1=2
2^2=4
2^3=8
2^4=6
2^5=2 .. the trend continues 2,4,8,6

97^275 - 32^44

Effectively 7^275 - 2^44
7^275 will have the same last digit as 7^3
2^44 will have the same last digit as 2^4

7^3-2^4

3 - 6
(Same as 13 - 6 carried over from the tens digfit)
= 7

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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)
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Quote:
What is the digit on the units place in the expanded value of 97275–324497275–3244?

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

in no. 97, cyclicity of unit's digit 7 is 7,9,3,1
and 275 /4 gives 3 as a remainder.
in no.32, cyclicity of unit's digit 2 is 2,4,8,6
and 44/4 gives 0 as remainder.

hence the equation is ...3-...6 which will end up to be 7 as the unit's digit.
hence option D
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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
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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.
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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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97^275 - 32^44

Following 7's cyclicity of 4 -

7^1 = 7 = ending in 7
7^2 = 49 = ending in 9
7^3 = 343 = ending in 3
7^4 = 2401 = ending in 1

So, 97^275 will be like

7^275 since we are dealing only with digit in units place currently.

Hence,

7^275 is of the type 7^3 and hence will end in a 3

Similarly, following 2's cyclicity of 4 -

2^1 = 2 = ending in 2
2^2 = 4 = ending in 4
2^3 = 8 = ending in 8
2^4 = 16 = ending in 6

So, 32^44 will be like

2^44 since we are dealing only with digit in units place currently.

Hence,

2^44 is of the type 2^4 and hence will end in a 6

Now we have 2 numbers each ending in 3 and 6 respectively.

A difference of the 2 numbers will leave a 7 in the units digit of the resulting number.
Example: 153 - 46 = 107

Hence answer is 7 and option D
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Cyclicity of digits.

Units digits :-
7^1 = 7
7^2 = 9 (49)
7^3 = 3 (343)
7^4 = 1 (2401)

As we can see, 7^4 has 1 at its units digit. Thus 7^5 will have a 7 at its units digit and the whole cycle will be repeated.

Now, 97 has 7 at its units place and the 7 will dictate the units digit of all its exponents, i.e. 97^1 has 7 at its units digit, 97^2 will have 9(9409) at its units digit and so on.
Thus 97^275 = [97^(272)]*[97^3]
272 = 4*68
Units digit of 97^272 = 1
Units digit of 97^3 = 3

Thus units digit of 97^275 = 3


Similarly,
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 6 (16)
2^5 = 2 (32)

So, units digit of 32^44 will be the same as the units digit of 2^44
Which is the same as the units digit of 2^4 = 6.

So the units digit of the given equation = 3 - 6 = 7.
Hence (D)


P.S. since 97^275 is clearly greater than 32^44, the resulting number when 32^44 is subtracted from 97^275 will be positive.
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This one is about cyclicity , so you have to consider cyclicity of the unit digits (7 and 2) in order to find the resultant unit digit:

cyclicity units of 7:

7^1=7
7^2=9
7^3=3
7^4=1

cyclicity units of 2:

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

So, for 97^275:

275=4K+3, so the unit number will have 3 as units

So, for 32^44:

44=4K, so the unit number will have 6 as units

Then, since 97^275 > 32^44, and the units of each are 3 and 6 respectively, the resultant unit digit must be 7.

D is the answer.
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Unit digit of 97^275 would be same as 7^275, which would be same as 7^3 (the unit digits of 7^x repeat after x = 4), which is 3.
Similarly, unit digit of 32^44 would be same as 2^44, which would be same as 2^4 (the unit digits of 2^x repeat after x = 4), which is 6.

Now, ____3-____6 will be same as 13-6 = 7 (where "____x" represents some number string ending with x)
Hence, answer should be (D).
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Quote:
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

Unit digit of \(97^{275}–32^{44}\) = Unit digit of \(7^{275}–2^{44}\)

CONCEPT: The unit digit of any number depends on the unit digit of the number so any digit other than unit digit is irrelevant such as \(97^{275}\) will have same unit digit as \(387^{275}\) or \(7^{275}\)

Now, Unit digit of \(7^{275}–2^{44}\) = Unit digit of \(7^3–2^4\)

CONCEPT: The cyclicity of unit digit of 7 and 2 is 4 i.e.e unit digits repeat after every 4th power. therefore Remainder (275/4) = 3 hence unit digit of \(7^275\) = unit digit of \(7^3\)

Unit digit of \(7^3–2^4\) = 3-6 = 7

Answer: Option D
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We are looking at the unit digits so we just need to consider the unit digit of each number.
So we need to look at
7^275 - 2^44
Now, unit digits of
7^1 = 7
7^2 = 9
7^3 = 3
7^4 = 1
7^5 = 7
So, we see after every 4th power the unit digit starts repeating.
We essentially need
7^3 as 275 = 4*68 + 3
Similarly
44 is a multiple of 4 so we will look at unit digit of 2^4
The problem is reduced to 3 - 6.
As 3 is a small number than 6 and is present at the units place, it will take a carry over from the tens place.
So, it becomes 13 - 6 = 7
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Bunuel
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 question was provided by Experts Global
for the Heroes of Timers Competition
In unit digit calculation only Unit digit of the number matters and any digit of base number is irrelevant

Unit of \(97^{275} – 32^{44}\) = Unit of \(7^{275} – 2^{44}\) = Unit of \(7^{272+3} – 2^{44}\) = Unit of \(7^{3} – 2^{4}\) = 3-6 = 13-6 = 7
Answer: Option D

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