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Posts: 32
Updated on: 18 Feb 2019, 04:28
Originally posted by devmillenium2k on 17 Nov 2014, 21:27.
Last edited by Bunuel on 18 Feb 2019, 04:28, edited 1 time in total.
Last edited by Bunuel on 18 Feb 2019, 04:28, edited 1 time in total.
Updated.
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I found this post in some other forum.
Let's divide this method into three parts and we will discuss each part one by one:
a. Last two digits of numbers which end in one
b. Last two digits of numbers which end in 3, 7 and 9
c. Last two digits of numbers which end in 2,4,6,8
Last two digits of numbers ending in 1
Let's start with an example.
Find the last two digits of \(41^{2789\)}
Multiply the tens digit of the number (4 here) with the last digit of the exponent (9 here) to get the tens digit. The unit digit will always equal to one.
In no time at all you can calculate the answer to be 61 (4 × 9 = 36). Therefore, 6 will be the tens digit and one will be the unit digit
Here are some more examples:
What are the last two digits of \(31^{786}\)?
Last two digits will be 81 (3 × 6 gives 8 as tens digit and 1 as unit digit)
Find the last two digits of \(71^{56747}\)
Last two digits will be 91 (7 × 7 gives 9 as tens digit and 1 as unit digit)
Now try to get the answer to this question within 10 seconds:
Find the last two digits of \(51^{456}\) × \(61^{567}\)
The last two digits of \(51^{456}\) will be 01 and the last two digits of \(61^{567}\) will be 21. Therefore, the last two digits of \(51^{456}\) × \(61^{567}\) will be the last two digits of 01 ×21 = 21
Last two digits of numbers ending in 3, 7 or 9
Find the last two digits of \(19^{266}\).
\(19^{266}\) = \((19^{2})^{133}\). Now, \(19^{2}\) ends in 61 (\(19^{2}\) = 361) therefore, we need to find the last two digits of \((61)^{133}\).
Once the number is ending in 1 we can straight away get the last two digits with the help of the previous method. The last two digits are 81 (6 × 3 = 18, so the tens digit will be 8 and last digit will be 1)
Find the last two digits of \(33^{288}\).
\(33^{288}\) = \((33^{4})^{72}\). Now \(33^{4}\) ends in 21 (\(33^{4}\) = \(33^{2}\) × \(33^{2}\) = 1089 × 1089 = xxxxx21) therefore, we need to find the last two digits of \(21^{72}\). By the previous method, the last two digits of \(21^{72}\) = 41 (tens digit = 2 × 2 = 4, unit digit = 1)
So here's the rule for finding the last two digits of numbers ending in 3, 7 and 9:
Convert the number till the number gives 1 as the last digit and then find the last two digits according to the previous method
Now try the method with a number ending in 7:
Find the last two digits of \(87^{474}\).
\(87^{474}\) = \(87^{472}\) ×\(87^{2}\) = \((87^{4})^{118}\) × \(87^{2}\) = \((69 × 69)^{118}\) × 69 (The last two digits of \(87^{2}\) are 69) =\(61^{118}\) × 69 = 81 × 69 = 89
Last two digits of numbers ending in 2, 4, 6 or 8
There is only one even two-digit number which always ends in itself (last two digits) i.e. 76 raised to any power gives the last two digits as 76. Therefore, our purpose is to get 76 as last two digits for even numbers. We know that \(24^{2}\) ends in 76 and \(2^{10}\) ends in 24. Also, 24 raised to an even power always ends with 76 and 24 raised to an odd power always ends with 24. Therefore, \(24^{34}\) will end in 76 and \(24^{53}\) will end in 24.
Let's apply this funda:
Find the last two digits of \(2^{543}\).
\(2^{543}\) = \((2^{10})^{54}\) × \(2^{3}\) = \((24)^{54}\) (24 raised to an even power) × \(2^{3}\) = 76 × 8 = 08
(NOTE: Here if you need to multiply 76 with \(2^{n}\), then you can straightaway write the last two digits of\(2^{n}\) because when 76 is multiplied with \(2^{n}\) the last two digits remain the same as the last two digits of \(2^{n}\). Therefore, the last two digits of 76 × \(2^{7}\) will be the last two digits of \(2^{7}\) = 28)
Same method we can use for any number which is of the form 2^{n}. Here is an example:
Find the last two digits of \(64^{236}\).
\(64^{236}\) = \((2^{6})^{236}\) =\(2^{1416}\) = \((2^{10})^{141}\) × \(2^{6}\) = \(24^{141}\) (24 raised to odd power) × 64 = 24 × 64 = 36
Here are some more examples:
Find the last two digits of \(62^{586}\).
\(62^{586}\) = \((2×31)^{586}\) = \(2^{586}\)×\(3^{586}\) = \((2^{10})^{58}\)×\(2^{6}\) × \(31^{586}\) = 76 × 64 × 81 = 84
Find the last two digits of \(54^{380}\).
\(54^{380}\) = \((2×3^{3})^{380}\) = \(2^{380}\)×\(3^{1140}\) = \((2^{10})^{38}\) × \((3^{4})^{285}\) = 76 × \(81^{285}\)= 76 × 01 = 76.
Find the last two digits of \(56^{283}\).
\(56^{283}\) = \((2^{3}×7)^{283}\) = \(2^{849}\)×\(7^{283}\) = \((2^{10})^{84}\) × \(2^{9}\) × \((7^{4})^{70}\) × \(7^{3}\) = 76 × 12 × \((01)^{70}\) × 43 = 16
Find the last two digits of \(78^{379}\).
\(78^{379}\) = \((2×39)^{379}\) = \(2^{379}\) × \(39^{379}\) = \((2^{10})^{37}\) × \(2^{9}\) × \((39^{2})^{189}\) × 39 = 24 × 12 × 81 × 39 = 92
Let's divide this method into three parts and we will discuss each part one by one:
a. Last two digits of numbers which end in one
b. Last two digits of numbers which end in 3, 7 and 9
c. Last two digits of numbers which end in 2,4,6,8
Last two digits of numbers ending in 1
Let's start with an example.
Find the last two digits of \(41^{2789\)}
Multiply the tens digit of the number (4 here) with the last digit of the exponent (9 here) to get the tens digit. The unit digit will always equal to one.
In no time at all you can calculate the answer to be 61 (4 × 9 = 36). Therefore, 6 will be the tens digit and one will be the unit digit
Here are some more examples:
What are the last two digits of \(31^{786}\)?
Last two digits will be 81 (3 × 6 gives 8 as tens digit and 1 as unit digit)
Find the last two digits of \(71^{56747}\)
Last two digits will be 91 (7 × 7 gives 9 as tens digit and 1 as unit digit)
Now try to get the answer to this question within 10 seconds:
Find the last two digits of \(51^{456}\) × \(61^{567}\)
The last two digits of \(51^{456}\) will be 01 and the last two digits of \(61^{567}\) will be 21. Therefore, the last two digits of \(51^{456}\) × \(61^{567}\) will be the last two digits of 01 ×21 = 21
Last two digits of numbers ending in 3, 7 or 9
Find the last two digits of \(19^{266}\).
\(19^{266}\) = \((19^{2})^{133}\). Now, \(19^{2}\) ends in 61 (\(19^{2}\) = 361) therefore, we need to find the last two digits of \((61)^{133}\).
Once the number is ending in 1 we can straight away get the last two digits with the help of the previous method. The last two digits are 81 (6 × 3 = 18, so the tens digit will be 8 and last digit will be 1)
Find the last two digits of \(33^{288}\).
\(33^{288}\) = \((33^{4})^{72}\). Now \(33^{4}\) ends in 21 (\(33^{4}\) = \(33^{2}\) × \(33^{2}\) = 1089 × 1089 = xxxxx21) therefore, we need to find the last two digits of \(21^{72}\). By the previous method, the last two digits of \(21^{72}\) = 41 (tens digit = 2 × 2 = 4, unit digit = 1)
So here's the rule for finding the last two digits of numbers ending in 3, 7 and 9:
Convert the number till the number gives 1 as the last digit and then find the last two digits according to the previous method
Now try the method with a number ending in 7:
Find the last two digits of \(87^{474}\).
\(87^{474}\) = \(87^{472}\) ×\(87^{2}\) = \((87^{4})^{118}\) × \(87^{2}\) = \((69 × 69)^{118}\) × 69 (The last two digits of \(87^{2}\) are 69) =\(61^{118}\) × 69 = 81 × 69 = 89
Last two digits of numbers ending in 2, 4, 6 or 8
There is only one even two-digit number which always ends in itself (last two digits) i.e. 76 raised to any power gives the last two digits as 76. Therefore, our purpose is to get 76 as last two digits for even numbers. We know that \(24^{2}\) ends in 76 and \(2^{10}\) ends in 24. Also, 24 raised to an even power always ends with 76 and 24 raised to an odd power always ends with 24. Therefore, \(24^{34}\) will end in 76 and \(24^{53}\) will end in 24.
Let's apply this funda:
Find the last two digits of \(2^{543}\).
\(2^{543}\) = \((2^{10})^{54}\) × \(2^{3}\) = \((24)^{54}\) (24 raised to an even power) × \(2^{3}\) = 76 × 8 = 08
(NOTE: Here if you need to multiply 76 with \(2^{n}\), then you can straightaway write the last two digits of\(2^{n}\) because when 76 is multiplied with \(2^{n}\) the last two digits remain the same as the last two digits of \(2^{n}\). Therefore, the last two digits of 76 × \(2^{7}\) will be the last two digits of \(2^{7}\) = 28)
Same method we can use for any number which is of the form 2^{n}. Here is an example:
Find the last two digits of \(64^{236}\).
\(64^{236}\) = \((2^{6})^{236}\) =\(2^{1416}\) = \((2^{10})^{141}\) × \(2^{6}\) = \(24^{141}\) (24 raised to odd power) × 64 = 24 × 64 = 36
Here are some more examples:
Find the last two digits of \(62^{586}\).
\(62^{586}\) = \((2×31)^{586}\) = \(2^{586}\)×\(3^{586}\) = \((2^{10})^{58}\)×\(2^{6}\) × \(31^{586}\) = 76 × 64 × 81 = 84
Find the last two digits of \(54^{380}\).
\(54^{380}\) = \((2×3^{3})^{380}\) = \(2^{380}\)×\(3^{1140}\) = \((2^{10})^{38}\) × \((3^{4})^{285}\) = 76 × \(81^{285}\)= 76 × 01 = 76.
Find the last two digits of \(56^{283}\).
\(56^{283}\) = \((2^{3}×7)^{283}\) = \(2^{849}\)×\(7^{283}\) = \((2^{10})^{84}\) × \(2^{9}\) × \((7^{4})^{70}\) × \(7^{3}\) = 76 × 12 × \((01)^{70}\) × 43 = 16
Find the last two digits of \(78^{379}\).
\(78^{379}\) = \((2×39)^{379}\) = \(2^{379}\) × \(39^{379}\) = \((2^{10})^{37}\) × \(2^{9}\) × \((39^{2})^{189}\) × 39 = 24 × 12 × 81 × 39 = 92
Good work, but I want to point out that is well beyond the scope of GMAT. I have not seen any official GMAT problem so far that asked one to find the last two digits of a number. Yes, they do ask you to find the last digit of powers of numbers, but not the last two digits.
Dabral
Dabral
Posts: 1
Find last two digit of 37^78945
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BubbleBalloon
Joined: 21 Jun 2019
Last visit: 12 Nov 2024
Posts: 15
Location: Canada
Concentration: Operations, Organizational Behavior
Posts: 15
dabral
Posts: 1
mkmeena55444
57 is the answer
ag153
Joined: 16 Feb 2017
Last visit: 15 May 2022
Posts: 80
Location: India
Concentration: Finance, Strategy
Schools: XLRI Columbia CBS '24 Harvard '24 ISB '23 Stanford '24
GPA: 3.69
Schools: XLRI Columbia CBS '24 Harvard '24 ISB '23 Stanford '24
Posts: 80
How did you make 76*64*81= 84
And 87^2=69?
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And 87^2=69?
Posted from my mobile device
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