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14 Jul 2019, 23:44
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A function # is defined on positive integers as #(a) = #(a – 1) + 1. If the value of #(1) is 1, then what is the value of #(3) ?
A . 0
B. 1
C. 2
D. 3
E. 4
A . 0
B. 1
C. 2
D. 3
E. 4
vinayakvaish
Joined: 29 May 2018
Last visit: 19 Oct 2020
Posts: 93
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Location: India
Schools: Marshall '22 (A$) IIMA PGPX "21 (D) Duke '22 (D) IIMB EPGP "21 (A) McCombs '22 (A) IIMC MBAEx "21 (WD) Georgia Tech '22 (A$) Johnson '22 (WD)
GMAT 1: 700 Q48 V38
GMAT 2: 720 Q49 V39
GPA: 4
Schools: Marshall '22 (A$) IIMA PGPX "21 (D) Duke '22 (D) IIMB EPGP "21 (A) McCombs '22 (A) IIMC MBAEx "21 (WD) Georgia Tech '22 (A$) Johnson '22 (WD)
GMAT 2: 720 Q49 V39
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14 Jul 2019, 23:54
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Lets use f instead of #
f(3)= f(2)+1
f(2)= f(1)+1
f(1)=1
substituting values
f(3)= f(1)+1+1
f(3)=3
Ans is 3
Hit Kudos!!
f(3)= f(2)+1
f(2)= f(1)+1
f(1)=1
substituting values
f(3)= f(1)+1+1
f(3)=3
Ans is 3
Hit Kudos!!
14 Jul 2019, 23:55
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To simplify the question we can write:
Given: #(a) = #(a – 1) + 1
f(a) = f(a – 1) + 1
Now f(1)=1,
f(2)=1+1=2
f(3)=2+1=3
So #(3)=3
Hence D
Posted from my mobile device
Given: #(a) = #(a – 1) + 1
f(a) = f(a – 1) + 1
Now f(1)=1,
f(2)=1+1=2
f(3)=2+1=3
So #(3)=3
Hence D
Posted from my mobile device
