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Find the last two digits of any no

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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
d. Last two digits of numbers which end in 4, 6 and 8

Before we start, let me mention binomial theorem in brief as we will need it for our calculations.


cat 2007 cat 2008 xat 2008 mba 2008 last two digits

Last two digits of numbers ending in 1

Let’s start with an example.

What are the last two digits of 31786?

Solution: 31786 = (30 + 1)786 = 786C0 x 1786 + 786C1 x 1785 x (30) + 786C2 x 1784 x 302 + …, Note that all the terms after the second term will end in two or more zeroes. The first two terms are 786C0 x 1786 and 786C1 x 1785 x (30). Now, the second term will end with one zero and the tens digit of the second term will be the product of 786 and 3 i.e. 8. Therefore, the last two digits of the second term will be 80. The last digit of the first term is 1. So the last two digits of 31786 are 81.

Now, here is the shortcut:


cat 2007 cat 2008 mba 2008 xat 2008 finding the last two digits
Here are some more examples:


Find the last two digits of 412789


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 units digit)


Find the last two digits of 7156747


Last two digits will be 91 (7 × 7 gives 9 and 1 as units digit)


Now try to get the answer to this question within 10 s:


Find the last two digits of 51456 x 61567


The last two digits of 51456 will be 01 and the last two digits of 61567 will be 21. Therefore, the last two digits of 51456 x 61567 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 19266.


19266 = (192)133. Now, 192 ends in 61 (192 = 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 x 3 = 18, so the tens digit will be 8 and last digit will be 1)


Find the last two digits of 33288.


33288 = (334)72. Now 334 ends in 21 (334 = 332 x 332 = 1089 x 1089 = xxxxx21) therefore, we need to find the last two digits of 2172. By the previous method, the last two digits of 2172 = 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:

cat 2007 cat 2008 xat 2008 mba 2008 last two digits

Now try the method with a number ending in 7:

Find the last two digits of 87474.

87474 = 87472 x 872 = (874)118 x 872 = (69 x 69)118 x 69 (The last two digits of 872 are 69) = 61118 x 69 = 81 x 69 = 89

If you understood the method then try your hands on these questions:

Find the last two digits of:

1. 27456
2. 7983
3. 583512

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) - 76 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 242 ends in 76 and 210 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, 2434 will end in 76 and 2453 will end in 24.

Let’s apply this funda:

Find the last two digits of 2543.

2543 = (210)54 x 23 = (24)54 (24 raised to an even power) x 23 = 76 x 8 = 08

(NOTE: Here if you need to multiply 76 with 2n, then you can straightaway write the last two digits of 2n because when 76 is multiplied with 2n the last two digits remain the same as the last two digits of 2n. Therefore, the last two digits of 76 x 27 will be the last two digits of 27 = 28)

Same method we can use for any number which is of the form 2n. Here is an example:

Find the last two digits of 64236.

64236 = (26)236 = 21416 = (210)141 x 26 = 24141 (24 raised to odd power) x 64 = 24 x 64 = 36

Now those numbers which are not in the form of 2n can be broken down into the form 2n ´ odd number. We can find the last two digits of both the parts separately.

Here are some examples:

Find the last two digits of 62586.

62586 = (2 x 31)586 = 2586 x 3586 = (210)58 x 26 x 31586 = 76 x 64 x 81 = 84

Find the last two digits of 54380.

54380 = (2 x 33)380 = 2380 x 31140 = (210)38 x (34)285 = 76 x 81285 = 76 x 01 = 76.

Find the last two digits of 56283.

56283 = (23 x 7)283 = 2849 x 7283 = (210)84 x 29 x (74)70 x 73 = 76 x 12 x (01)70 x 43 = 16

Find the last two digits of 78379.

78379 = (2 x 39)379 = 2379 x 39379 = (210)37 x 29 x (392)189 x 39 = 24 x 12 x 81 x 39 = 92



24 Comments
    quantques
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    quantquesSat, 01 Feb 2014 21:17:03 -0000

    how to find the last two digits of 347^347?? plz help sumone

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    subha_gym
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    subha_gymThu, 12 Sep 2013 19:10:20 -0000

    please can anyone help me with this problem??
    Find the last two digits of : 153763519717
    ans is 35 but give me the explanation

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    c m punk
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    c m punkTue, 03 Sep 2013 15:19:40 -0000

    appreciated a lot

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    doesntmatter
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    doesntmatterMon, 19 Sep 2011 09:42:06 -0000

    the source of information must be mentioned in reference..
    The snippets are an exact copy of an article of TotalGadha.com
    Breach of intellectual property.. sad to see this !

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    priswa
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    priswaSat, 09 Oct 2010 05:54:25 -0000

    what do you mean by last 2 digits of no? Is it any place value?

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    debrey
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    debreyThu, 29 Apr 2010 06:38:31 -0000

    Can you pls show me solution on how to find the last two digit of this number 680^237?

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    beedistinct
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    beedistinctFri, 16 Jul 2010 09:13:57 -0000

    The last two digits will be 00. (68 × 10)^237 = 68^237 X 10^237
    In fact its last 237 digits will be zeroes.

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    grimreaper
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    praveenThu, 04 Mar 2010 15:26:52 -0000

    another method wld be to see the cyclic nature of tens place… for any power of a no…. try it out………..and notice the pattern

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    grimreaper
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    praveenThu, 04 Mar 2010 15:04:10 -0000

    hmmm…….. does anyone realise why this happens ;) think abt it… just learning a trick aint enough…… see the actual reason behind it………

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    a123nu
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    a123nuSat, 31 Oct 2009 19:59:47 -0000

    how to find the last two digits of ..13 × 45x 56 x78 x34x 67×89×23x12 how to find wen power is not raised?

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    sharada venkat
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    sharada venkatFri, 21 May 2010 03:18:23 -0000

    here since theres a 5 and 6 (or any other even number in 1's place) the ones digit will be 0. we jus have to multiply the digits at 1's place in the rest of the digits to get the 10's place.
    the ans is 60.

    paritosh_ieheco
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    paritosh_iehecoWed, 09 Nov 2011 14:14:46 -0000

    sir, i understand that how that 0 came at 1s place but please elaborate on how that 6 came….

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    swastikdash
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    swastikdashSat, 05 Sep 2009 12:25:27 -0000

    i may sound silly but what bout last digit ending in 5??

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    deja-vu
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    deja-vuWed, 30 Sep 2009 06:23:10 -0000
    Reply to This
    raj_citm
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    RajTue, 01 Sep 2009 11:19:26 -0000

    thank you sir

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    nidhithatsme
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    nidhi gargThu, 27 Aug 2009 19:29:37 -0000

    @ clark

    6 is not equal to 3^2

    6^12 = ( 2 * 3 ) ^12 = 2^10 * 2^2 * 3^12
    ..24 * 4 * 9^6
    96 * 81^3
    ..96 * ..41
    = ..36

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    LKP DREAMERZZZ
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    1 Aim IIMFri, 21 Aug 2009 10:39:58 -0000

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    prithvi10089
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    Prithvi Raj AmbatiMon, 12 Oct 2009 10:34:02 -0000

    That's a wonderful trick ! Thanks a lot !

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    iflex
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    amrit madhavFri, 24 Jul 2009 10:23:47 -0000

    excellent, actually i was searching for this kind of tricks only so as to crack this kind of problems.

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    vinaycat
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    VINAY KUMAR NWed, 22 Jul 2009 11:32:36 -0000

    Superb Concept… thanks a lot

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    clarke
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    sylvester clarkeMon, 08 Jun 2009 22:18:36 -0000

    hi there…i think the method is not full proof..
    using the above method how would we find the last 2 digits of the following no.

    6^12

    =(3*2)^12 = 3^12 * 2^12 = 81^3 * 2^10 * 2^2 =41 *76*04 = 64 but the answer is 36

    other abberations are 126^24 etc

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    vinaycat
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    VINAY KUMAR NWed, 22 Jul 2009 11:44:33 -0000

    hi clarke,
    The method works for any number,
    6^12 = (3 × 2)^12
    = 3^12 × 2^10 × 2^2
    = (3^4)^3 × 24 × 4
    = 81^3 × 24 × 4
    = 41 × 24 × 4
    = **36 hence the method works
    note: i have considered only last two digits in the calculations above.

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    userdce
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    user dceSat, 30 May 2009 07:52:17 -0000

    highly appreciated…very valuable

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    igreen
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    mayank sangwanWed, 21 Jan 2009 20:41:22 -0000

    I am very much thankful to you sir for providing such a miraculous techniques.

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    bonnynit
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    bonnynitWed, 21 Jan 2009 12:45:46 -0000
    wow its amazing

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    Sanjs
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    sanjeeveesTue, 13 Jan 2009 20:16:28 -0000

    i appreciate the concept and steps can be further decreased by applying the negative modulo concept for e.g. 87^474= -13^474 = (169)^237=(-31)^237=3X7 and 1= -11= 100-11=89
    Thank you (Sanj)

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    bhati
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    bhatiMon, 29 Dec 2008 11:01:46 -0000

    thanks appreciated

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    ronakdangi
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    ronakdangiTue, 21 Oct 2008 11:52:21 -0000

    its really grt………….

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    Shravan
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    M Shravan KumarFri, 17 Oct 2008 22:47:25 -0000

    thank u very much sir

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    shaky20
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    shaky20Tue, 14 Oct 2008 09:10:53 -0000

    AWESOME TRICK SIR

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Last Updated At Dec 07, 2012
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