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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # In multiplying two positive integers a and b, Ron reversed the digits

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Senior Manager  P
Status: Manager
Joined: 02 Nov 2018
Posts: 280
In multiplying two positive integers a and b, Ron reversed the digits  [#permalink]

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Difficulty:   25% (medium)

Question Stats: 83% (02:14) correct 17% (01:43) wrong based on 18 sessions

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In multiplying two positive integers a and b, Ron reversed the digits of the two-digit number a. His errorneous product was 161. What is the correct value of the product of a and b?

(A) 116

(B) 161

(C) 204

(D) 214

(E) 224
Manager  S
Joined: 19 Jan 2019
Posts: 110
Re: In multiplying two positive integers a and b, Ron reversed the digits  [#permalink]

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1
Let's say the two digits of integer a be xy
Integer b be d..

As per the question we need to find (xy)(d)

By reversing the digits
(10y + X)* d = 161..
(10y + X)* d = 23*7
D = 7
10y + X = 23.. the only possible solution for this equation is X = 3 and y = 2...( Single digits)
How??? Because 10y is a multiple of 10... And X can only take single digit values....

Originally a = xy = 32..
a*b = xy*d = 32*7 = 224...

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Intern  B
Joined: 30 May 2017
Posts: 17
Re: In multiplying two positive integers a and b, Ron reversed the digits  [#permalink]

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reversed digits of a multiplied by b=161

161 = 23*7 so if the reversed digits of a are 23, then a=32 Re: In multiplying two positive integers a and b, Ron reversed the digits   [#permalink] 21 Mar 2019, 08:01
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# In multiplying two positive integers a and b, Ron reversed the digits  