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Question

Which of the integers 10, 11, 12 and 13 can be written as the sum of squares of four integers (allowing repetition)?

The correct answer is

All

Integers as Sum of Four Squares

The question asks which of the integers 10, 11, 12, and 13 can be written as the sum of the squares of four integers, allowing repetition. This is a classic problem in number theory related to representing numbers as sums of squares.

According to Lagrange's four-square theorem, every natural number can be represented as the sum of four integer squares. This means that all integers, including 10, 11, 12, and 13, should be expressible in this form.

Let's demonstrate how each of these integers can be written as the sum of four squares:

10 as Sum of Four Squares

We need to find four integers a, b, c, and d such that $a^2 + b^2 + c^2 + d^2 = 10$. We can use squares of small integers like $0^2=0$, $1^2=1$, $2^2=4$, $3^2=9$, etc.

One way to express 10 as the sum of four squares is:

  • $\text{10} = 3^2 + 1^2 + 0^2 + 0^2 = 9 + 1 + 0 + 0$

Since we found a combination, 10 can be written as the sum of four squares.

11 as Sum of Four Squares

Next, we check if 11 can be written as the sum of squares of four integers. Using squares 0, 1, 4, 9:

We look for a combination $a^2 + b^2 + c^2 + d^2 = 11$. Here is one possibility:

  • $\text{11} = 3^2 + 1^2 + 1^2 + 0^2 = 9 + 1 + 1 + 0$

Thus, 11 can also be written as the sum of four squares.

12 as Sum of Four Squares

Now, let's examine the integer 12. Can it be written as the sum of squares of four integers?

We need $a^2 + b^2 + c^2 + d^2 = 12$. Here is one way:

  • $\text{12} = 2^2 + 2^2 + 2^2 + 0^2 = 4 + 4 + 4 + 0$

Another possible way is:

  • $\text{12} = 3^2 + 1^2 + 1^2 + 1^2 = 9 + 1 + 1 + 1$

Since we found ways to express 12 in this form, 12 can be written as the sum of four squares.

13 as Sum of Four Squares

Finally, let's check the integer 13. Can 13 be written as the sum of squares of four integers?

We need $a^2 + b^2 + c^2 + d^2 = 13$. Consider this combination:

  • $\text{13} = 3^2 + 2^2 + 0^2 + 0^2 = 9 + 4 + 0 + 0$

This combination works, confirming that 13 can be written as the sum of four squares.

Sum of Squares Conclusion

We have shown examples for each integer:

  • 10 = $3^2 + 1^2 + 0^2 + 0^2$
  • 11 = $3^2 + 1^2 + 1^2 + 0^2$
  • 12 = $2^2 + 2^2 + 2^2 + 0^2$
  • 13 = $3^2 + 2^2 + 0^2 + 0^2$

Therefore, all the integers 10, 11, 12, and 13 can be written as the sum of squares of four integers.

The option that states 'All' is correct because all the given integers can be represented in this way.

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Important Questions from Integers

  1. A two digit number is five times the sum of its digits. If 9 is added to the number, the digits interchange their positions. The sum of digits of the number is :
  2. If 6 times the larger number is divided by the smaller number, we get 7 as quotient and 7 as remainder.  Also, if nine times the smaller number is divided by the larger number, we get 7 as the quotient and 21 as the remainder. What are the two numbers?

  3. Which one of the following is the greatest number by which the product of three consecutive even numbers would be exactly divisible ?
  4. What is the natural number n for which $3^9 + 3^{12} + 3^{15} + 3^n$ is a perfect cube of an integer ?
  5. The least integer when multiplied by 2940 becomes a perfect square is
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