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Bunuel
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If n is an integer greater than 2 and t = 10^n − 132, if t is repres 17 Dec 2019, 01:26
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69% (01:55) correct 31% (01:52) wrong based on 396 sessions

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If n is an integer greater than 2 and \(t = 10^n − 132\), if t is represented in decimal notation, which can be the sum of all digits of t?

A. 30
B. 31
C. 32
D. 33
E. 34

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If n is an integer greater than 2 and t = 10^n − 132, if t is repres 05 Sep 2022, 04:06
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Bunuel
If n is an integer greater than 2 and \(t = 10^n − 132\), if t is represented in decimal notation, which can be the sum of all digits of t?

A. 30
B. 31
C. 32
D. 33
E. 34

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\(10^n\), where n is a non-negative integer, has \(n+1\) digits: 1, followed by \(n\) zeros. For example:
    \(10^0=1\). 1 has one digit: 1, followed by 0 zeros;
    \(10^1=10\). 10 has two digits: 1, followed by 1 zeros;
    \(10^2=100\). 100 has four digits: 1, followed by 2 zeros;
    \(10^3=1,000\). 1,000 has four digits: 1, followed by 3 zeros;
    ...

\(10^n-132\) will have \(n\) digits: \(n-3\) 9's, followed by 868. For example:
    If n = 3, then \(10^n − 132=868\). The sum of the digits = 22;
    If n = 4, then \(10^n − 132=9,868\). The sum of the digits = 31;
    If n = 5, then \(10^n − 132=99,868\). The sum of the digits = 40;
    If n = 6, then \(10^n − 132=999,868\). The sum of the digits = 49;
    ...

As we can see, the answer is 31, B.

Answer: B.

If you can deduce the above right away, then you could get the formula for the sum of the digits of \(10^n − 132\) and solve from there. \(10^n − 132\) will have \(n-3\) 9's, followed by 868:

    \(9*(n - 3) + 8 + 6 + 8 = 9n -5\).


So, the answer must be 5 less than a multiple of 9. Only B works: 31 = 4*9 - 5.

Hope it's clear.

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Hope it helps.
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If n is an integer greater than 2 and t = 10^n − 132, if t is repres 17 Dec 2019, 02:15
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Method 1-

\(10^n\)= 1 mod 9, for n>2
132= 6 mod 9

\(10^n − 132\)= (9+1-6) mod 9= 4 mod 9

Only 31 among options gives 4 remainder when divided by 9

Method 2

t=\(10^n-132\)= 999....(n-3) times....868

Sum of digits
= 9+9+9..(n-3)times+8+6+8
= 9(n-3)+22

when n=4, sum of digits = 31
when n=5, sum of digits= 40

B


Bunuel
If n is an integer greater than 2 and \(t = 10^n − 132\), if t is represented in decimal notation, which can be the sum of all digits of t?

A. 30
B. 31
C. 32
D. 33
E. 34

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If n is an integer greater than 2 and t = 10^n − 132, if t is repres 10 Jan 2020, 09:29
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nick1816

I think you made a typo... some of digits should follow the pattern: S = 22 + 9 x (n-1)..


Thank you for showing the arithmetic progression though
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If n is an integer greater than 2 and t = 10^n − 132, if t is repres 10 Jan 2020, 10:37
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When n=3, t = 1000-132 = 868 (number of 9's = n-3 = 3-3 = 0)
When n=4, t = 10000-132 = 9868 (number of 9's = n-3 = 4-3 = 1)
When n=5, t = 100000-132 = 99868 (number of 9's = n-3 = 5-3 = 2)

and so on....

Sum= 9(n-3)+8+6+8

Camach700
nick1816

I think you made a typo... some of digits should follow the pattern: S = 22 + 9 x (n-1)..


Thank you for showing the arithmetic progression though
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If n is an integer greater than 2 and t = 10^n − 132, if t is repres 07 Sep 2024, 08:50
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Hi Bunuel,

I understood your explanation, however I am still curious to understand what decimal representation is being talked about in the question?
''t is represented in decimal notation''
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If n is an integer greater than 2 and t = 10^n − 132, if t is repres 07 Sep 2024, 09:26
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divyasaxena2825
Hi Bunuel,

I understood your explanation, however I am still curious to understand what decimal representation is being talked about in the question?
''t is represented in decimal notation''
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­
Based 10 notation, or decimal notation, is just a way of writing a number using 10 digits: 1, 2, 3, 4, 5, 6, 7, 8, and 0 (usual way), in contrast, for example, to binary numeral system (base-2 number system) notation.
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If n is an integer greater than 2 and t = 10^n − 132, if t is repres 23 Sep 2025, 12:57
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