To solve last digit problems

Type # 1:

questions of the form…

2003 x 2004 x 4235161006 x 432657178001 x 42315098002 x 423087004

concept:

Last n digits of any product depends on the product of last n digits. so just multiply last n digits of each term…find the product, take last n digits of the product n multiply it with the next term…continue this for all terms.

suppose v had to find last digit of the product above…

so multiply last digits.

3 x 4 = 12. dont worry abt 1 in 12. just remember 2 and multiply it with next no. 2 x 6 = 12. so 2, 2 x 1= 2, 2 x 2 = 4, 4 x 4 = 16.
so, the ans is 6.

for last 2 digits:

03 x 04 x 06 x 01 x 02 x 04 = 76

for last 3 digits:

003 x 004 x 006 x 001 x 002 x 004 = 576.

trust me, last 4 digits wont be asked…as then it becomes bulky…questions in cat are tricky…

Type # 2:

of the form…last digits of 432^43567. i.e base ^ power

i know most of us know this concept…wud take it concisely…

look for the variation in last digit of higher powers of last digit of the base…i.e. 2 here.

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

so we can say that 2,4,8,6 will keep repeating…no. of different digits that the last digits of higher powers can take is known as cyclicity. every digit has a cyclicity.

to find last digit, find last digit of :

(last digit of base) ^ (power % cyclicity of last digit)

when, power % cyclicity of last digit = 0,

take,

(last digit of base) ^ (cyclicity of last digit)

in the above example…

432^43567.

2^ (43567 % 4) = 2^3 = 8 answer.

with little practice….u can do all such questions mentally.

Type # 3

Find last 2 digits of 34291^201.

when the last digit is 1,9,0,5…try finding a pattern in last 2 digits…u’ll get one…then solve the question accordingly…

last 2 digits of 91^1 = 91
last 2 digits of 91^2 = 81
last 2 digits of 91^3 = 71
last 2 digits of 91^4 = 61
.
.
.
.
.
.
.
last 2 digits of 91^10 = 01
last 2 digits of 91^11 = 91

see, we again got 91 as last 2 digits…so v can say that the cyclicity of 91 for last 2 digits is 11 -1 = 10

so the answer shud be 91.

lets take one more example…

find last 2 digits of
(49)^(37)^(38 )^(39)…(3700)

see…

last 2 digits of 49^1 => 49
last 2 digits of 49^2 => 01
last 2 digits of 49^3 => 49
last 2 digits of 49^4 => 01.

can v say that for all odd powers, answer wud be 01?
yes!

since the power of 49 is odd…the answer shud be 01.

what if the question was

(49)^(37)^(38 )^(39)…(3700)%100 = ?
answer wud be 01%100 = 1

(49)^(37)^(38 )^(39)…(3700) % 20 = ?
answer wus be 01 % 20 = 1.
bcoz, to c remainder with 20, v need last 2 digits only.

(49)^(37 )^(38 )^(39)…(3700) % 10
=> 49 % 10 = 9

this method might become tedious when the last 2 digits are unfriendly…like 37, 82 etc. but i have neva seen such figures appearing in cat…to solve such figures, v need euler’s or binomial…ill be taking it when v discuss remainders…today’s questions are based on the above concept only..

Questions for today..

1. last digit of 3677^400 – 689^84

2. last digit of 11^11 + 12^12….1000^1000.

3. Last 2 digits of : 233527 x 54725 x 64535379 x 64536247 x 63546342 x 435272599 x 7454453279 x 63546337 x 543624334547 x 74547459 x 7454373 x 6545347359 x 735473451…do i need to say….its tricky…not lengthy!!

4. no. of zeroes at the end in 1^1×2^2×3^3x…250^250.

5. no. of zeroes at the end in 1! x 2! x 3! x…25!

6. Last non-zero digit of 25!

7. Last non-zero digit of 1! x 2! x 3!…15!

Find last 2 digits of…

8. 81^(371)^(372)^…(400)

9. 11 ^ (25)^(31)^(41)…(1001)

10. 7 ^ 2501.

11. 3^2537837.

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19 Responses to “To solve last digit problems”

  1. ekta Says:

    Hi Bhavin,

    Nice explanation/post. Do you also post the anwers somewhere?
    If not can yuo send me teh answers for this excercise.

    Cheers,
    Ekta

  2. meetbhavin Says:

    REGRETS for not putting the solutions…

    1. last digit of 3677^400 – 689^84—- 0

    2. last digit of 11^11 + 12^12….1000^1000
    Ans 3

    from 1^1 till 10^10, we’ll get 1,6,3,6,5,6,7,4,9,0. the order will change but the set will be of these 10 digits only…just an observation!!

    so, we’re adding the bove sum (1000-10 ) /10 = 99 times…so the last digit shud be last digit of 47 x 99 = 3.

    3. Last 2 digits of : 233527 x 54725 x 64535379 x 64536247 x 63546342 x 435272599 x 7454453279 x 63546337 x 543624334547 x 74547459 x 7454373 x 6545347359 x 735473451
    Ans – 50

    there’s one 25 n one 2…that product will make the last 2 digits 50. now 50 x any odd no. will always leave 50 as last 2 digits…n all the other nos. are odd

    4. Last non-zero digit of 25!
    Ans – 5

    keep multiplying last digits…not a matter of more than 90 secs.
    5s… 5 ( 1+ 2+….50) = 6375
    25s…25(1+2+..10) = 1375
    125s…125(1+2) = 375

    total = 8125

    5. no. of zeroes at the end in 1! x 2! x 3! x…25!—-56

    5-9…5 nos. one 0 each = 5
    10-14…5 nos. two 0 each=10
    15-19…5 nos. three 0 each=15
    20-24…5 nos. four 0 each=20
    25…1 no. with 6 0s = 6

    total = 56

    6. Last non-zero digit of 25!
    Ans – 4

    7. Last non-zero digit of 1! x 2! x 3!…15!
    ans 6

    same approach as above

    Find last 2 digits of…

    8. 81^(371)^(372)^…(400)—81

    9. 11 ^ (25)^(31)^(41)…(1001)—-51

    10. 7 ^ 2501.—-07

    11. 3^2537837.—-63

    Eular No. for 100 is 40, so
    3^2537837 mod 100 —> 3^(-3) mod 100
    =>(((3*33)^(-3))*(33^3))%100=> ((-1)(33^3))%100=>-37 (mod 100)
    -37 (mod 100)=>63 (mod 100)

  3. Amit Damahe Says:

    hi……….. can u help me for finding last non zero digit of n! here n can be any 2 or 3 digit number.

  4. rules Says:

    hi if a number is of the format a^b where a is not an integer then how will u calculate the last m digits before the decimal point

  5. sasa_10 Says:

    i want to know the last 10 digits in the series
    1^1 + 2^2 + …………………+1000^1000;

    THANKS FOR YOUR HELP;

  6. Souvik Says:

    Found this very helpful. Thanks!

  7. Last digit problems - TestMagic Forums Says:

    [...] Last digit problems Guys, found a very interesting article on problems involving last digits of very large numbers..check this out…it has got good examples too..hope you find it helpful. The same blog contains other useful articles too. To solve last digit problems Welcome to Bhavin’s Blog [...]

  8. shreeom Says:

    Can any one give the solutions with full explanation.
    Thanks in advance……

  9. Cistina Says:

    Hi can you help me finding the last digit of the power 2^1001 thanks ………..

  10. shiv Says:

    pl tell me how to find the last digit of 13 raised to the power 2003

  11. Ikosarakt Says:

    This digit is 7, if you consider the sequence of 13^1, 13^2, 13^3, you find that all 13^3, 13^7, 13^x (where x+1 div 4 = x+1/4) have 7 in the last digit.

  12. nagarathinam Says:

    I WANT EXPLANATION FOR LAST DIGIT OF 38^2011 IS 2


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