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Algebra Level 4

{ A = 1 × 2 + 3 × 4 + 5 × 6 + … + 37 × 38 + 39 B = 1 + 2 × 3 + 4 × 5 + … + 36 × 37 + 38 × 39 \begin{cases} {A = 1\times 2 + 3\times4 + 5\times 6 + \ldots+37\times38 + 39 } \\ { B = 1 + 2\times 3 + 4 \times 5 + \ldots + 36 \times 37 + 38 \times 39 } \end{cases}

The values of A A and B B are obtained by writing multiplcation and addition operators in an alternating pattern between successive integers as described above. Find ∣ A − B ∣ |A-B| .

Notation: ∣ ⋅ ∣ | \cdot | denotes the absolute value function .


The answer is 722.

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Ravi Dwivedi by Ravi Dwivedi , 24, India

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2 solutions

Ravi Dwivedi
Jul 3, 2015

Therefore

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This appears much simpler if you write it in summation notation. We have,

A − 39 = ∑ k = 1 19 ( 2 k ) ( 2 k − 1 ) and B − 1 = ∑ k = 1 19 ( 2 k ) ( 2 k + 1 ) A-39=\sum_{k=1}^{19}(2k)(2k-1)\quad\textrm{and}\quad B-1=\sum_{k=1}^{19}(2k)(2k+1)

Subtract the first equation from the second to get,

( B − A ) + 38 = ∑ k = 1 19 ( 2 k ) ( 2 k + 1 − 2 k + 1 ) ( B − A ) + 38 = 4 ∑ k = 1 19 k = 4 × 19 × 20 2 = 760 B − A = 760 − 38 = 722 (B-A)+38=\sum_{k=1}^{19}(2k)(2k+1-2k+1)\\ (B-A)+38=4\sum_{k=1}^{19}k=4\times\frac{19\times 20}{2}=760\\ B-A=760-38=722

Hence, for this problem, we have,

∣ A − B ∣ = B − A = 722 |A-B|=B-A=\boxed{722}

In this problem, in the second line of calculation, we used the sum of first k k positive integers formula which is ∑ k = 1 n k = n ( n + 1 ) 2 \sum\limits_{k=1}^n k=\frac{n(n+1)}2 .

Prasun Biswas - 5 years, 11 months ago

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Thanks for an alternative!!

Ravi Dwivedi - 5 years, 11 months ago

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It's not an alternative solution. It's the same solution as yours but presented in a formal way using summation notation.

Prasun Biswas - 5 years, 11 months ago

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@Prasun Biswas – Yeah...same thing.....!!

Ravi Dwivedi - 5 years, 11 months ago

Exactly same solution

Kushagra Sahni - 5 years, 11 months ago
Chew-Seong Cheong
Oct 25, 2019

A = 0 + 1 × 2 + 3 × 4 + 5 × 6 + ⋯ × 36 + 37 × 38 + 39 B = 1 + 2 × 3 + 4 × 5 + 6 × 7 + ⋯ + 37 + 38 × 39 ⟹ B − A = 1 + 2 × 2 + 2 × 4 + 2 × 6 + ⋯ + 2 × 38 − 39 = 1 + 4 ( 1 + 2 + 3 + ⋯ + 19 ) − 39 = 1 + 4 × 19 × 20 2 − 39 = 722 \begin{aligned} A & = 0 + 1 \times \blue 2 + 3 \times \blue 4 + 5 \times \blue 6 + \cdots \times \blue{36} + 37 \times \blue {38} + 39 \\ B & = 1 + \blue 2 \times 3 + \blue 4 \times 5 + \blue 6 \times 7 + \cdots + 37 + \blue{38} \times 39 \\ \implies B-A & = 1 + 2 \times \blue 2 + 2 \times \blue 4 + 2 \times \blue 6 + \cdots + 2 \times \blue {38} - 39 \\ & = 1 + 4(1+2+3+ \cdots + 19) - 39 \\ & = 1 + 4 \times \frac {19\times 20}2 - 39 \\ & = \boxed{722} \end{aligned}

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