Aptitude Discussion

Q. |
The largest number amongst the following that will perfectly divide $101^{100} – 1$ is: |

✖ A. |
$100$ |

✔ B. |
$10,000$ |

✖ C. |
$100^{100}$ |

✖ D. |
$100,000$ |

**Solution:**

Option(**B**) is correct

The easiest way to solve such problems for objective exam purposes is trial and error or by back substituting answers in the choices given.

$101^2 = 10,201$.

$101^2 - 1 = 10,200$. This is divisible by 100.

Similarly try for $101^3 - 1 = 1,030,301 - 1 = 1,030,300$.

So you can safely conclude that $(101^1 - 1)$ to $(101^9 - 1)$ will be divisible by 100.

$(101^{10} - 1)$ to $(101^{99} - 1)$ will be divisible by 1000.

Therefore, $(101^{100} - 1)$ will be divisible by **10,000.**

**Edit:** For an alternative solution using binomial expansion, check comment by **Anuj Mittal.**

**Deepak**

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**Abhinav**

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i understand

$101^2 - 1 = 10,200$. This is divisible by 100.

So you can safely conclude that $(101^10 - 1)$ to $(101^100 - 1)$ will be divisible by 1000.

please explain

Therefore, $(101^{100} - 1)$ will be divisible by 10,000

**Soumik**

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I think it will be 100. As $101^{100} -1 = (100+1)^{100} -1$

$=(100^{100}+^{100}C_1 100^{99} + ...+1-1$

$= 100$ (.....)

Soumik you are right, but question says LARGEST number. One of the option choices clearly mentions $10,000$ as a choice. Which if true will indicate that is has to be divisible by $100$ as well.

**Anuj Mittal**

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we can also do this question by binomial theorem.

consider $101^100=(100+1)^100$ and apply binomial expansion the last two no.s in the expansion we get are 10000 and 1 and all the other no.s in the expansion are greater than 10000 and are also multiple of 10000 , so the last 5 digits we get are 10001 and subtracting 1 from it gives 10000.

we have a formula (1+x)^n = 1+nx. So 101^100=(1+100)^100-1=1+(100*100)-1=10^4 which is 10000