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28 Mar 2025, 00:30
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Difficulty:
35%
(medium)
Question Stats:
77% (02:21) correct
23%
(02:14)
wrong
based on 81
sessions
History
Date
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What is the sum of integers from 1 to 100 that are divisible by 2 or 5 ?
A. 550
B. 1050
C. 2550
D. 3050
E. 5050
A. 550
B. 1050
C. 2550
D. 3050
E. 5050
28 Mar 2025, 00:32
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Step 1: Identify numbers divisible by 2 or 5 in the range [1,100].
- Numbers divisible by 2: 2, 4, 6, ..., 100 (Arithmetic sequence)
First term (a) = 2, Common difference (d) = 2, Last term = 100
Number of terms (n) = \(\frac{100 - 2}{2} + 1 = 50\)
Sum = \(\frac{50}{2} \times (2 + 100) = 2550\)
- Numbers divisible by 5: 5, 10, 15, ..., 100
First term (a) = 5, Common difference (d) = 5, Last term = 100
Number of terms (n) = \(\frac{100 - 5}{5} + 1 = 20\)
Sum = \(\frac{20}{2} \times (5 + 100) = 1050\)
Step 2: Avoid double-counting numbers divisible by both 2 and 5 (i.e., 10).
- Numbers divisible by 10: 10, 20, 30, ..., 100
First term (a) = 10, Common difference (d) = 10, Last term = 100
Number of terms (n) = \(\frac{100 - 10}{10} + 1 = 10\)
Sum = \(\frac{10}{2} \times (10 + 100) = 550\)
Step 3: Apply Inclusion-Exclusion Principle
Total sum = (Sum of multiples of 2) + (Sum of multiples of 5) - (Sum of multiples of 10)
\(2550 + 1050 - 550 = 3050\)
Final Answer: 3050
- Numbers divisible by 2: 2, 4, 6, ..., 100 (Arithmetic sequence)
First term (a) = 2, Common difference (d) = 2, Last term = 100
Number of terms (n) = \(\frac{100 - 2}{2} + 1 = 50\)
Sum = \(\frac{50}{2} \times (2 + 100) = 2550\)
- Numbers divisible by 5: 5, 10, 15, ..., 100
First term (a) = 5, Common difference (d) = 5, Last term = 100
Number of terms (n) = \(\frac{100 - 5}{5} + 1 = 20\)
Sum = \(\frac{20}{2} \times (5 + 100) = 1050\)
Step 2: Avoid double-counting numbers divisible by both 2 and 5 (i.e., 10).
- Numbers divisible by 10: 10, 20, 30, ..., 100
First term (a) = 10, Common difference (d) = 10, Last term = 100
Number of terms (n) = \(\frac{100 - 10}{10} + 1 = 10\)
Sum = \(\frac{10}{2} \times (10 + 100) = 550\)
Step 3: Apply Inclusion-Exclusion Principle
Total sum = (Sum of multiples of 2) + (Sum of multiples of 5) - (Sum of multiples of 10)
\(2550 + 1050 - 550 = 3050\)
Final Answer: 3050
28 Mar 2025, 00:43
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Integers divisible by 2 = 2,4,...100
Sum = 2*50*51/2 = 2550
Integers divisible by 5 = 5,10,...100
Sum = 5*20*21/2 = 1050
Sum total = 2550+1050 = 3600
This however double counts integer divisible by both 2 and 5. We need to eliminate that
Integers divisible by 10 = 10,20,...100
Sum = 10*10*11/2 = 550
Total sum = 3600-550 = 3050
Sum = 2*50*51/2 = 2550
Integers divisible by 5 = 5,10,...100
Sum = 5*20*21/2 = 1050
Sum total = 2550+1050 = 3600
This however double counts integer divisible by both 2 and 5. We need to eliminate that
Integers divisible by 10 = 10,20,...100
Sum = 10*10*11/2 = 550
Total sum = 3600-550 = 3050
Bunuel
