A Nice Olympiads Simplification Trick | No Calculator Allowed

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[@rleroygordon]

One quick step with 8 * (2500 - 50) is to do the distributive operation:
8 * (2500 - 50) =
8 * 2500 - 8 * 50 =
20000 - 400 =
19600

[@Yurka-Shturka]

This is the sum of a geometric progression

[@AbhijitBhattacharyaju]

It’s sum of geometric progression so
S=a(r^n-1/r-1)=7*(7^5-1)/(7-1)=(7/6)*(49*49*7-1)=(7/6)*(7*(2500-100+1)-1)
=(7/6)*(7*2401-1)
=(7/6)*(16807-1)
=(7/6)*(16806)
=7*2801=19607

[@fadetoblah2883]

I factored the original expression like so:
7³(7²+7+1+[1/7]+[1/7²]), and then proceeded with the following, fairly simple calculations...
= 343(57+[56/343])
= 19551+56
= 19607

[@sie_khoentjoeng4886]

I think we can calculate
8 * 49 * 50 + 7
= 8 * 50 * 49 + 7
= 400 * 49 + 7
= 19600 + 7
= 19607

[@ManojGupta-bp5ws]

Bilkul sahi. For easy calculation we should always multiply number having factor 5 with even number first.he. Is making calculation difficult

[@129140163]

I came up with 19607 in my head
7^5 = 16807
7^4 = 2401, 16807 + 2401 = 19208
7^3 = 343, 19208 + 343 = 19551
7^2 = 49, 19551 + 49 = 19600
7^1 = 7, 19600 + 7 = 19607

[@als2cents679]

8 * 49 * 50 + 7
= 400 * 49 + 7
= 400 * 50 - 400 + 7
= 20000 - 400 + 7
= 19607

[@TheChongsam]

Starting from right to left, 7+49+49*7=343,..343*7,add these terms

[@charlesmitchell5841]

Ok, good to know in case the calculator companies go out of business 😅.

[@jerrypaquette5470]

I came up with the right answer with a lot less math.

[@alestee4241]

8*49*50=4*49*100=(200-4)*100

[@user-hm3js8js8y]

8* 49*50→ 4*2*49*50→4*49*100→196*100

[@rabotaakk-nw9nm]

3:40 8•49•50+7=49•2•2•50•2+7=
=98•2•100+7=196•100+7=19607 😁

[@-datolith2775]

😀

[@heathertilton3056]

7⁵+7⁴+7³+7²+7¹=19,607

[@SpamtonSpamton-pm6hi]

interesting, but so much slower than just summing the powers of 7

[@rleroygordon]

The problem is with trying to figure out powers of 7 in your head.

[@SpamtonSpamton-pm6hi]

@@rleroygordon yeah, but it just seems like using a lawnmower to cut a beard. by the time you figure out that relationship, you could've already calculated 7^4 as 2401 due to easy multiplication with 49^2. maybe if it was a sum with the largest term 7^8 or 7^10 i could understand. (and even then i'd use the finite geometric series formula instead as it uses way less operations than the other methods)

[@benjaminchang1382]

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