# Remainders and the Binomial Theorem ## 5 thoughts on “Remainders and the Binomial Theorem”

1. Alia |

Hi, I’m finding this method a little confusing. I understood the binomial theorem but am not able to extend my understanding to the remainder concept. I hope you can clear a couple of doubts
1. For the question on 23^111 divided by 121. Why have we used the 22+1 format and not the 11k+1 like we used for the 17^1717 divided by both 8 & 9?
2. How do we come to the conclusion of the format being 8m+1 (17^1717 divided by 8), what does the 8m signify? Does it refer to the part of the expansion that will give the remainder as 0 as it is readily divisible by 8?
3. When using the concept of binomial theorem in remainders how much of the expansion do we need too concentrate on in order get the remainder?

2. seema |

for last question, can you plz explain the last step, as how 2*99*7 when divided by 49 yields 2*1*7 or 14?

3. catcracker |

2 / 49 leaves 2, 99 / 49 leaves 1 and 7 / 49 leaves 7. Remainder of product = product of remainders as we saw in an earlier post…

regards
J

4. kronos |

why to write the term 49m(redundant)… got confused because of it.

• catcracker |

Because otherwise other readers would end up wondering where it vanished! 🙂 The point is that there is some term there, but it is divisible by 49 and hence will not contribute to the remainder.

regards
J