Math Problem Statement

Kids math problem. A chocolate shop is closing down and needs to sell all their chocolates. Your Mum only allows you to buy enough to fit in her handbag. The chocolates were counted in groups of 10 but there were 3 left over. The chocolates were counted in groups of 2 but there was 1 left over. How many chocolates could you buy? How many possible answers can you come up with?

Solution

This problem involves finding a number of chocolates that fits the conditions when grouped in 10s and 2s. Let’s break this down.

Problem Statement:

  • Groups of 10: There are 3 chocolates left over when counted in groups of 10. Mathematically, this is written as: N3 (mod 10)N \equiv 3 \ (\text{mod} \ 10) This means the total number of chocolates NN gives a remainder of 3 when divided by 10.

  • Groups of 2: There is 1 chocolate left over when counted in groups of 2. Mathematically, this is written as: N1 (mod 2)N \equiv 1 \ (\text{mod} \ 2) This means NN gives a remainder of 1 when divided by 2.

Finding the possible values of NN:

We need to find a number that satisfies both conditions.

  1. From N3 (mod 10)N \equiv 3 \ (\text{mod} \ 10), we know NN can be expressed as: N=10k+3N = 10k + 3 for some integer kk.

  2. Now, substitute this into the second condition, N1 (mod 2)N \equiv 1 \ (\text{mod} \ 2): 10k+31 (mod 2)10k + 3 \equiv 1 \ (\text{mod} \ 2) Simplifying: (10kmod2)+(3mod2)1 (mod 2)(10k \mod 2) + (3 \mod 2) \equiv 1 \ (\text{mod} \ 2) Since 10kmod2=010k \mod 2 = 0 (because 10 is even) and 3mod2=13 \mod 2 = 1, we get: 0+11 (mod 2)0 + 1 \equiv 1 \ (\text{mod} \ 2) This is true, so no further conditions are needed.

Conclusion:

The possible values of NN are all numbers of the form N=10k+3N = 10k + 3, where kk is an integer. The sequence of possible values is: N=3,13,23,33,43,53,63,73,83,93,103,113,N = 3, 13, 23, 33, 43, 53, 63, 73, 83, 93, 103, 113, \dots

These numbers are spaced by 10 units, and there are infinitely many possible solutions. The number of chocolates you can buy depends on how much space is available in the handbag.


Would you like more details or have any questions about this solution?

Here are 5 related questions to deepen your understanding:

  1. What does it mean for a number to be congruent modulo another number?
  2. Can you find another method to solve for NN given two modular conditions?
  3. How would the solution change if there were 2 chocolates left over when counted in groups of 5?
  4. Could we solve this problem with only the condition N3 (mod 10)N \equiv 3 \ (\text{mod} \ 10)?
  5. What is the smallest NN such that both conditions hold and N>100N > 100?

Tip: In problems with remainders like this, look for patterns in the multiples and solve them using modular arithmetic.

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Math Problem Analysis

Mathematical Concepts

Modular Arithmetic
Number Theory

Formulas

N ≡ 3 (mod 10)
N ≡ 1 (mod 2)

Theorems

Chinese Remainder Theorem

Suitable Grade Level

Grades 6-8