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# What least number must be subtracted from 1936

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Senior Manager
Joined: 10 Apr 2018
Posts: 267
Location: United States (NC)
What least number must be subtracted from 1936  [#permalink]

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11 Apr 2019, 11:34
2
00:00

Difficulty:

35% (medium)

Question Stats:

64% (02:09) correct 36% (01:57) wrong based on 28 sessions

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What least number must be subtracted from 1936 so that the remainder when divided by 9,10,15 in each case is 7.

(A) 39
(B)44
(C)51
(D)129
(E)141

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~You Just Can't beat the person who never gives up~ Babe Ruth
Intern
Joined: 31 May 2018
Posts: 13
Location: India
GPA: 4
Re: What least number must be subtracted from 1936  [#permalink]

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11 Apr 2019, 13:36
Step I-

LCM of 9, 10, 15 = 90

Step II-

$$\frac{1936}{90}$$ = quotient (21), remainder (46)

To get 7 as a remainder:
46-x = 7
x= 39

The least number is 39 that should be subtracted from 1936.

i.e. 1936-39 = 1897

[*] $$\frac{1897}{9}$$ = quotient (210), remainder (7)

[*] $$\frac{1897}{10}$$ = quotient (189), remainder (7)

[*] $$\frac{1897}{15}$$ = quotient (126), remainder (7)

Re: What least number must be subtracted from 1936   [#permalink] 11 Apr 2019, 13:36
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