1. 有一袋彈珠,每2顆一數,每3顆一數或每7顆一數都剛好數完,這袋彈珠至少有幾顆? 2. 265公車每15分鐘開出一班,245公車每隔20分鐘開出一班,現在兩班公車同時開出,下一次兩班公車又同時開出是多久... 1. 有一袋彈珠,每2顆一數,每3顆一數或每7顆一數都剛好數完,這袋彈珠至少有幾顆? 2. 265公車每15分鐘開出一班,245公車每隔20分鐘開出一班,現在兩班公車同時開出,下一次兩班公車又同時開出是多久之後? 3. 將39顆巧克力和52片餅乾,平均分給小朋友,每人分到的巧克力顆數要一樣多,餅乾也要一樣多,要分給最多位小朋友,每人會分到幾顆巧克力和幾片餅乾? 4. 1789 最少要減去多少,才能被19整除? 5. 1456最少要添加多少,才能被27整除?

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Understand the Problem

This is a set of math questions related to number theory and arithmetic. These questions involve finding the Least Common Multiple (LCM), Greatest Common Divisor (GCD), and remainders after division.

Answer

2. 42 3. 60 4. 3 chocolates and 4 cookies 5. 3 6. 2
Answer for screen readers
  1. 42
  2. 60
  3. 3 chocolates and 4 cookies
  4. 3
  5. 2

Steps to Solve

Here are the solutions to the math problems:

  1. Find the LCM of 2, 3, and 7

The problem states that the number of marbles is divisible by 2, 3, and 7. We need to find the least common multiple (LCM) of these numbers to find the minimum number of marbles. Since 2, 3, and 7 are all prime numbers, their LCM is simply their product. $LCM(2, 3, 7) = 2 \times 3 \times 7 = 42$

  1. Find the LCM of 15 and 20

The problem asks when the two buses will depart at the same time again. We need to find the least common multiple (LCM) of 15 and 20. Prime factorization of 15: $15 = 3 \times 5$ Prime factorization of 20: $20 = 2 \times 2 \times 5 = 2^2 \times 5$ $LCM(15, 20) = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60$

  1. Find the GCD of 39 and 52, then divide

The problem requires dividing 39 chocolates and 52 cookies equally among the maximum number of children. This means finding the greatest common divisor (GCD) of 39 and 52. Prime factorization of 39: $39 = 3 \times 13$ Prime factorization of 52: $52 = 2 \times 2 \times 13 = 2^2 \times 13$ $GCD(39, 52) = 13$ Each child gets $39 \div 13 = 3$ chocolates and $52 \div 13 = 4$ cookies.

  1. Find the remainder of 1789 divided by 19

The problem asks what is the smallest number we should subtract from 1789, so that the result is divisible by 19. We should divide 1789 by 19 to get a remainder. $1789 \div 19 = 94$ with a remainder of $3$. Therefore, we need to subtract 3.

  1. Find the remainder of 1456 divided by 27

The problem asks what is the smallest number to add to 1456 such that the result is divisible by 27. To solve this, first find the remainder when 1456 is divided by 27. Then subtract the remainder from 27. $1456 \div 27 = 53$ with a remainder of $25$. $27 - 25 = 2$ Therefore, we need to add 2.

  1. 42
  2. 60
  3. 3 chocolates and 4 cookies
  4. 3
  5. 2

More Information

These problems cover basic number theory concepts that are foundational for more advanced math. Understanding LCM, GCD, and remainders is very important.

Tips

  • Forgetting to find the least common multiple or greatest common divisor.
  • Making calculation errors in division or factorization.
  • Misinterpreting the question, for instance, adding instead of subtracting in questions 5 and 6, or vice versa.

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