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61. The total number of 3 digit numbers which have two or more consecutive digits identical is:

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Solution:
In each set of 100 numbers, there are 10 numbers whose tens digit and unit digit are same. Again in the same set there are 10 numbers whose hundreds and tens digits are same. But one number in each set of 100 numbers whose Hundreds, Tens and Unit digit are same as 111, 222, 333, 444 etc
Hence, there are exactly (10 + 10 - 1) = 19 numbers in each set of 100 numbers. Further there are 9 such sets of numbers
Therefore such total numbers = 19 × 9 = 171

Alternatively,
9 × 10 × 10 - 9 × 9 × 9 = 900 - 729 = 171
62. A magazine publisher publishes a monthly magazine of 84 pages. One I found that in a magazine 4 pages was missing. One out of them was page number 29 it is known that the page number of the last page of the magazine is 84, (including the cover pages). The numbers printed on the missing pages were :

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Solution:
Since, the magazine has 84 pages it means that there are 21 sheets of paper, which are folded in the middle. The pattern of pages will be this way,
Left Right
1, 2 83, 84
3, 4 81, 82
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-
-
-
-
-
-
-
29, 30 55, 56
63. The remainder when 6666..∞ is divided by 10 is :

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Solution:
Since,610→remainderis66610→remainderis666610→remainderis6666610→remainderis66666610→remainderis6andsoon.Thus,inallthesuchcasesTheremainderwillalwaysbe6
64. [(888×888×888)−(222×222×222)][(888×888)+(888×222)+(222×222)]=?

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Solution:
[(888×888×888)−(222×222×222)][(888×888)+(888×222)+(222×222)]=(8883−2223)[8882+(888×222)+2222]=888−222=666
65. The remainder of 3232327:

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Solution:
323232means32(32.32.32.32.......32times)andtheremainderof327is4So,4(32.32.32.32.......32times)74(2.2.2.2.2........32times)7Remainder=4Since,47→Remainder4427→Remainder2437→Remainder1447→Remainder4
66. The Remainder of 888222+2228883is:

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Solution:
Remainder of 888222+2228883     = 0
Since, 888 and 222 both (bases) are divisible by 3
67. Find the remainder when
10 + 102 + 103 + 104 + ........ + 1099 is divided by 6.

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Solution:
The remainder when 10 is divided by 6 is 4.
The remainder when 102 is divided by 6 is 4.
The remainder when 103 is divided by 6 is 4.
The remainder when 104 is divided by 6 is 4.
Thus, remainder is always 4.
So, the required remainder,
=4+4+4+4+.....99 times6=3966
Thus, Required remainder is 0.
68. The remainder of (3)67! divided by 80 is :

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Solution:
Since, 34=81  gives remainder 1 on divided by 80
So, 34n80  gives remainder 1
Thus, 367!80  will also give the remainder as 1
Since, 67! = 4n for a positive integer n
69. The remainder of 3997!40 is :

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Solution:
Since, ana+1   gives remainder 1 when 'n' is even.
Now since, 97! is an even number so remainder will be 1.
Note:-Factorial of any no. is even.
70. The remainder of 259255 is:

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Solution:
259canbeexpressedas28nSo,Remainder259255=Remainder28n255=Remainder256n255=1Requiredremainder=1

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