Number Play Class 6 Long Question Answer

Number Play Class 6 Long Question Answer

Number Play Class 6 Long Question Answer

Question 1.

Start with the number 1949. How many rounds will you take to reach the kaprekar constant with the digits of this numbers?

Solution:

Here, A = 9941 and B = 1499. So, C= 9941 - 1499 = 7442. ......(i)

New, A = 7442 and B = 2447. So, C = 7442 - 2447 = 4995. .....(ii)

New, A 9954 and B = 4599. So, C = 9954 - 4599 = 5355. ......(iii)

New, A = 5553 and B = 3555. So, C = 5553 - 3555 = 1998 ..........(iv)

New, A = 9981 and B = 1899. So, C = 9981 - 1899 = 8082. ........(v)

New, A = 8820 and B = 0288. So, C = 8820 - 0288 = 8532. ......(vi)

New, A = 8532 and B = 2358. So, C = 8532 -2358 = 6174. ..........(vii)

So, we reach the Kaprekar constant after 7 rounds.

Question 2.

Starting with the number 100, find how many steps you will have to take to reach 1 as per Collatz Conjecture?

Solution:

[latex]\frac{100}{2}[/latex] = 50, [latex]\frac{50}{2}[/latex] = 25, 25 × 3+1 = 76,

[latex]\frac{76}{2}[/latex] = 38, [latex]\frac{38}{2}[/latex] = 19, 19 × 3 + 1 = 58, [latex]\frac{58}{2}[/latex] = 29,

29 × 3 + 1 = 88, [latex]\frac{88}{2}[/latex] = 44, [latex]\frac{44}{2}[/latex] = 22, [latex]\frac{22}{2}[/latex] = 11,

11 × 3 + 1 = 34, [latex]\frac{34}{2}[/latex] = 17, 17 × 3 + 1 = [latex]\frac{52}{2}[/latex] = 26,

[latex]\frac{26}{2}[/latex] = 13, 13 × 3 + 1 = 40, [latex]\frac{40}{2}[/latex] = 20, [latex]\frac{20}{2}[/latex] = 10,

[latex]\frac{10}{2}[/latex] = 5, 5 × 3 + 1 = 16, [latex]\frac{16}{2}[/latex] = 8, [latex]\frac{8}{2}[/latex] = 4, [latex]\frac{4}{2}[/latex] = 2,

[latex]\frac{2}{2}[/latex] = 1.

Thus, we reach 1 after 20 steps.