AMC 8 Daily Practice - Calculation Tricks by Grouping

Complete problem set with solutions and individual problem pages

Problem 1 Easy

What is the value of (7 + 8 + 9) - 7 + (7 + 8 + 9) - 8 + (7 + 8 + 9) - 9?

  • A.

    -24

  • B.

    24

  • C.

    48

  • D.

    72

  • E.

    96

Answer:C

Through observation, we notice that the expression (7 + 8 + 9) appears repeatedly in the original formula.

The expression can be rewritten as: (7 + 8 + 9) - 7 + (7 + 8 + 9) - 8 + (7 + 8 + 9) - 9 = (7 + 8 + 9) + (7 + 8 + 9) + (7 + 8 + 9) - 7 - 8 - 9 = 3 \times (7 + 8 + 9) - (7 + 8 + 9) = (7 + 8 + 9) \times (3 - 1) = (7 + 8 + 9) \times 2 = 24 \times 2 = 48

Thus, the final result is \boxed{48}.

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Problem 2 Easy

What is the value of 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + 9 - 10 + 11 - 12 + 13 - 14 + 15 - 16 + \dots + 99 - 100?

  • A.

    5050

  • B.

    -51

  • C.

    -50

  • D.

    -5050

  • E.

    50

Answer:C

We group the addends in pairs: 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + 9 - 10 + 11 - 12 + 13 - 14 + 15 - 16 + \dots + 99 - 100 =(1 - 2) + (3 - 4) + (5 - 6) + (7 - 8) + (9 - 10) + (11 - 12) + (13 - 14) + (15 - 16) + \dots + (99 - 100) We find that when calculating each of the expressions within the parentheses, the result always  is  one, and there are fifty parentheses.  (1 - 2) + (3 - 4) + (5 - 6) + (7 - 8) + (9 - 10) + (11 - 12) + (13 - 14) + (15 - 16) + \dots + (99 - 100)=(-1) \times 50

So the result of the original expression is -50.

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Problem 3 Medium

What is the value of 169 - 168 - 167 + 166 + 165 - 164 - 163 + 162 + \dots + 9?

  • A.

    0

  • B.

    -9

  • C.

    5

  • D.

    9

  • E.

    84

Answer:D

By observing the arithmetic expression, we notice that the sum of the first four numbers is 0: 169 - 168 - 167 + 166 = 0

Examining the next set of four numbers, we find their sum is also 0: 165 - 164 - 163 + 162 = 0

Thus, the original expression can be rewritten as: 169 - 168 - 167 + 166 + 165 - 164 - 163 + 162 + \dots + 9 = (169 - 168 - 167 + 166) + (165 - 164 - 163 + 162) + \dots + 9

The problem now reduces to determining how many such groups exist. From 169 to 9, there are a total of 161 numbers.

When divided into groups of four, we get: 161 \div 4 = 40 groups with a remainder of 1.

Therefore, the original expression can be represented as: (169 - 168 - 167 + 166) + (165 - 164 - 163 + 162) + \dots + 9= 40 \times 0 + 9 = 9

The final result is \boxed{9}.

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Problem 4 Easy

What is the value of 2025 - 2024 + 2023 - 2022 + 2021 - 2020 + 2019 - 2018 + \dots + 3 - 2 + 1?

  • A.

    1012

  • B.

    1013

  • C.

    1014

  • D.

    1015

  • E.

    1016

Answer:B

By observing the arithmetic sequence, we notice that the sum of every two consecutive numbers from left to right is 1.

Therefore, we can group the numbers in pairs. Since there are 2025 numbers in total, dividing them into groups of two yields: 2025 \div 2 = 1012 groups with a remainder of 1.

Thus, the original expression can be rewritten as: 2025 - 2024 + 2023 - 2022 + 2021 - 2020 + 2019 - 2018 + \dots + 3 - 2 + 1 = (2025 - 2024) + (2023 - 2022) + (2021 - 2020) + (2019 - 2018) + \dots + (3 - 2) + 1 = \underbrace{1 + 1 + 1 + \dots + 1}_{1012 \text{ terms}} + 1 = 1 \times 1012 + 1 = 1013

The final result is \boxed{1013}.

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Problem 5 Easy

What is the value of 4 - 6 + 8 - 10 + \dots + 2020 - 2022 + 2024?

  • A.

    1014

  • B.

    1010

  • C.

    -1010

  • D.

    1016

  • E.

    1018

Answer:A

Through observation, we find that the sum of every two consecutive numbers is -2.

Grouping the sequence into pairs, we calculate the total number of terms as: 2024 \div 2 - 1 = 1011 numbers, 1011 \div 2 = 505\dots1 which forms 505 complete groups with 1 remaining term.

The original expression can thus be rewritten as: 4 - 6 + 8 - 10 + \dots + 2020 - 2022 + 2024 = (4 - 6) + (8 - 10) + \dots + (2020 - 2022) + 2024 = \underbrace{(-2) + (-2) + \dots + (-2)}_{505 \text{ terms}} + 2024 = (-2) \times 505 + 2024 = -1010 + 2024 = 1014.

The final result is \boxed{1014}.

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Problem 6 Medium

What is the value of 2+4-6+8+10-12+ \dots+92+94-96+98?

  • A.

    720

  • B.

    818

  • C.

    863

  • D.

    800

  • E.

    0

Answer:B

This problem is somewhat unique and requires careful observation. We notice that the operation signs repeat every three numbers in the pattern ++-.

Therefore, we can group every three numbers together. The sum of the first group is 0, the sum of the second group is 6, and the sum of the third group is 12, indicating that these sums form an arithmetic sequence with a common difference of 6.

Calculating the number of groups: 98 \div 2 = 49 terms, then 49 \div 3 = 16 groups with a remainder of 1 term. Thus, there are 16 complete groups.

We can rewrite the original expression as: 2 + 4 - 6 + 8 + 10 - 12 + \dots + 92 + 94 - 96 + 98 = (2 + 4 - 6) + (8 + 10 - 12) + \dots + (92 + 94 - 96) + 98 = 0 + 6 + 12 + \dots + 90 + 98

This forms an arithmetic sequence with: - First term a_1 = 0 - Common difference d = 6 - Last term a_{16} = 90 -

Applying the arithmetic series sum formula (Gauss's formula): S_n = \frac{n(a_1 + a_n)}{2} + \text{remaining term} S = \frac{16 \times (0 + 90)}{2} + 98 = 720 + 98 = 818

The final result is \boxed{818}.

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Problem 7 Easy

What is the value of 211\times 555+445\times 789+555\times 789+211\times 445?

  • A.

    999999

  • B.

    1000000

  • C.

    936840

  • D.

    1111111

  • E.

    100000

Answer:B

Method 1:

By observation, we notice that the expression contains common factors 211 and 789.

Applying the associative property of multiplication, the original expression can be rewritten as: 211 \times 555 + 445 \times 789 + 555 \times 789 + 211 \times 445 = 211 \times (555 + 445) + 789 \times (445 + 555) = 211 \times 1000 + 789 \times 1000 = (211 + 789) \times 1000 = 1000 \times 1000 = 1,000,000

Thus, the result is \boxed{1000000}.

Method 2:

By observation, we notice that the expression contains common factors 555 and 445.

Applying the associative property of multiplication, the original expression can be rewritten as: 211 \times 555 + 445 \times 789 + 555 \times 789 + 211 \times 445 = 555 \times (211 + 789) + 445 \times (789 + 211) = (555 + 445) \times (211 + 789) = 1000 \times 1000 = 1,000,000

Thus, the result is \boxed{1000000}.

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Problem 8 Medium

What is the value of 32\times 31-31\times 30+30\times 29-29\times 28+…+4\times 3-3\times 2+2\times 1?

  • A.

    992

  • B.

    0

  • C.

    1

  • D.

    257

  • E.

    512

Answer:E

By observing the first four terms, we notice the common factor 31.

Applying the associative property of multiplication, the first four terms can be rewritten as: 31 \times (32 - 30)

Similarly, analyzing terms five to eight, we identify the common factor 29 and rewrite them as: 29 \times (30 - 28)

The original expression can thus be transformed into a series of such grouped terms: 32 \times 31 - 31 \times 30 + 30 \times 29 - 29 \times 28 + \dots + 4 \times 3 - 3 \times 2 + 2 \times 1 = 31 \times (32 - 30) + 29 \times (30 - 28) + \dots + 3 \times (4 - 2) + 2 \times 1 = (31 + 29 + \dots + 3) \times 2 + 2

This forms an arithmetic sequence with: - First term a_1 = 3 - Last term a_n = 31 - Common difference d = 2 - Number of terms n = \frac{31 - 3}{2} + 1 = 15

Applying Gauss's formula for the sum of an arithmetic series: S = \frac{n}{2} \times (a_1 + a_n) S = \frac{15}{2} \times (3 + 31) = 255

Final calculation: 255 \times 2 + 2 = 510 + 2 = 512

The final result is \boxed{512}.

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