Question
How many positive integers ≤ 1260 are relatively prime to 1260?
Answer: Option D
:
D
Method 1: Conventional Approach
1260=22×32×5×7
We need to find the number of positive integers 1260 which are not divisible by 2, 3, 5 or 7.
Let A be the set consisting of multiples of 2≤1260;
B be the set consisting of multiples of 3;
C be the set consisting of multiples of 5 & D of multiples of 7.
A=12602=630,
B=12603=420,
C=12605=252,
D=12607=180
AΠB=12602×3=210AΠC=12602×5=126AΠD=12602×7=90
BΠC=12603×5=84
BΠD=12603×7=60CΠD=12605×7=36
AΠBΠC=12602×3×5=42
AΠBΠD=12602×3×7=30
AΠCΠD=12602×5×7=18
BΠCΠD=12603×5×7=12
AΠBΠCΠD=12602×3×5×7=6
So, A∪B∪C∪D=630+420+252+180−210−126−90−84−60−36+42+30+18+12−6=972
So, 1260 - 972 = 288 integers are relatively prime to 1260.
Method 2: Shortcut: Using the concept of Euler's number
The number of positive integers ≤1260 which are relatively prime to 1260 can be directly found by finding the Euler's number of 1260
=N(1−12)×(1−13)×(1−15)×(1−17)
=1260×12×23×45×67=288
(where 2,3,5,7 are the prime factors of 1260)
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:
D
Method 1: Conventional Approach
1260=22×32×5×7
We need to find the number of positive integers 1260 which are not divisible by 2, 3, 5 or 7.
Let A be the set consisting of multiples of 2≤1260;
B be the set consisting of multiples of 3;
C be the set consisting of multiples of 5 & D of multiples of 7.
A=12602=630,
B=12603=420,
C=12605=252,
D=12607=180
AΠB=12602×3=210AΠC=12602×5=126AΠD=12602×7=90
BΠC=12603×5=84
BΠD=12603×7=60CΠD=12605×7=36
AΠBΠC=12602×3×5=42
AΠBΠD=12602×3×7=30
AΠCΠD=12602×5×7=18
BΠCΠD=12603×5×7=12
AΠBΠCΠD=12602×3×5×7=6
So, A∪B∪C∪D=630+420+252+180−210−126−90−84−60−36+42+30+18+12−6=972
So, 1260 - 972 = 288 integers are relatively prime to 1260.
Method 2: Shortcut: Using the concept of Euler's number
The number of positive integers ≤1260 which are relatively prime to 1260 can be directly found by finding the Euler's number of 1260
=N(1−12)×(1−13)×(1−15)×(1−17)
=1260×12×23×45×67=288
(where 2,3,5,7 are the prime factors of 1260)
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