five digit multiplication?

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fruitmonster97

2492 posts

fruitmonster97

#1 h Nov 7, 2024, 4:41 PM

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What is the value of $9901\cdot101-99\cdot10101?$

$\textbf{(A) }2\qquad\textbf{(B) }20\qquad\textbf{(C) }21\qquad\textbf{(D) }200\qquad\textbf{(E) }2020$

This post has been edited 2 times. Last edited by fruitmonster97, Nov 7, 2024, 5:39 PM

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HonestCat

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HonestCat

#2 h Nov 7, 2024, 4:42 PM

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This is (A) 2

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clarkculus

236 posts

clarkculus

#3 h Nov 7, 2024, 4:43 PM • 3 Y

Y by centslordm, oinava, Silaban

I got a…

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hsuya1

174 posts

hsuya1

#4 h Nov 7, 2024, 4:43 PM • 1 Y

Y by Silaban

Yeah you can just look at the numbers mod 10 and just see that the last digit must be 2

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VigneshS32

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VigneshS32

#5 h Nov 7, 2024, 4:43 PM

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I got A, I bashed it.

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ethanzhang1001

1060 posts

ethanzhang1001

#6 h Nov 7, 2024, 4:44 PM

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this is actually kinda hard for #1 lol

the answer is A, notice $9901\cdot 101-99\cdot 10101=9900\cdot 101-99\cdot 10100+101-99$ and the left two multiplications are equal

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Countmath1

180 posts

Countmath1

#7 h Nov 7, 2024, 4:44 PM

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Write $9901= 10,000 - 99$ and $10101 = 10000 + 101$ , expand, simplify: $\textbf{(A)\ 2}.$

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Marcus_Zhang

980 posts

Marcus_Zhang

#8 h Nov 7, 2024, 4:44 PM • 1 Y

Y by TheEmperorPenguin

bruh spent 6 minutes doing this one and checking > : (

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fruitmonster97

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fruitmonster97

#9 h Nov 7, 2024, 4:44 PM

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We compute $9901\cdot101=1000001$ and $99\cdot10101=99999,$ so our answer is $1000001-999999=2,$ answer choice $\boxed{\textbf{(A)}}.$

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evt917

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evt917

#10 h Nov 7, 2024, 4:44 PM

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A so easy arithmetic

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bot1132

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bot1132

#11 h Nov 7, 2024, 4:47 PM

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(A) 2, just calculate

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alexanderhamilton124

397 posts

alexanderhamilton124

#12 h Nov 7, 2024, 4:47 PM • 1 Y

Y by Nobitasolvesproblems1979

i got 21 somehow :skull: don't question it

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ElaineGu

388 posts

ElaineGu

#13 h Nov 7, 2024, 4:51 PM • 1 Y

Y by YTM2010

Just look at last digits, 1-9 so it has to be 2

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pingpongmerrily

3635 posts

pingpongmerrily

#14 h Nov 7, 2024, 4:51 PM

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i did some weird substitution but just look and you can see it has to end with a 2 so A is the only choice

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mithu542

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mithu542

#15 h Nov 7, 2024, 4:53 PM

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pingpongmerrily wrote:

i did some weird substitution but just look and you can see it has to end with a 2 so A is the only choice

Option choice $(D)$ also ends in $2$ .

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xHypotenuse

778 posts

xHypotenuse

#36 h Nov 7, 2024, 6:20 PM

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SomeonecoolLovesMaths wrote:

A confirmed

I put 2 but why was my answer choice E for that one

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scrabbler94

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scrabbler94

#37 h Nov 7, 2024, 6:27 PM

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Modulo 10 kills this problem, since $9901 \cdot 101 - 99 \cdot 10101 \equiv 1 - 9 \equiv 2 \pmod{10}$ . Only (A) has units digit 2.

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wangzrpi

159 posts

wangzrpi

#38 h Nov 7, 2024, 6:35 PM

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ayush_agarwal wrote:

Yeah you can just look at the numbers mod 10 and just see that the last digit must be 2

Nice solution. I rounded the products and estimated

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TetraFish

1088 posts

TetraFish

#39 h Nov 7, 2024, 7:11 PM

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Pretty satisfying p1. $9901(101) - 99(9901 + 200) = 2(9901) - 99(200) = 2(9901) - 2(9900) = 2(1) = \boxed{2}$ , so A.

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SkatingKitty

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SkatingKitty

#40 h Nov 7, 2024, 7:52 PM

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One problem I got right

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akqbstr

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akqbstr

#41 h Nov 7, 2024, 7:53 PM

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Just look at the last digit.
9901*101, last digit is 1
99*10101, last digit 9
subtract them, last digit has to be 2

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YunboZ

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YunboZ

#42 h Nov 7, 2024, 10:24 PM

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A, I bashed it too

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gicyuraok2

1059 posts

gicyuraok2

#43 h Nov 7, 2024, 11:17 PM

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ooh new amc drop i guess i'll try all the problems

anyway free problem just compute and your good for A

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Existing_Human1

213 posts

Existing_Human1

#44 h Nov 7, 2024, 11:45 PM

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POV: You somehow fail to learn how to subtract and multiply, so you have to skip this problem. Come back to it, still forget how to multiply and subtract, and finally get it halfway through the test (I'm not actually that bad, just silly a bunch)

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