If six - digit number 5x2y6z is divisible by 7, 11 and 13, then the va

1.	If six - digit number 5x2y6z is divisible by 7, 11 and 13, then the value of (x – y + 3z) is:1. 92. 03. 74. 4
Answer» Correct Answer - Option 3 : 7 Ans: 3 Given: 5x2y6z is divisible by 7, 11 and 13 Concept used: 7 × 11 × 13 = 1001, when any 3 digit number multiplied by 1001 then it repeats itself and always completely divisible by 7,11 and 13. Calculation: Let 3 digit number be pqr. ⇒ pqr × 1001 = 5x2y6z ⇒ pqrpqr = 5x2y6z Compare both sides ⇒ p = 5, q = 6 and r = 2 Hence, that number is 562562. ⇒ 5x2y6z = 562562 by comparing both numbers, x = 6, y = 5, and z = 2 Required value = (x – y + 3z) ⇒ 6 – 5 + 3 × 2 ⇒ 7 ∴ The required value is 7.

If six - digit number 5x2y6z is divisible by 7, 11 and 13, then the value of (x – y + 3z) is:1. 92. 03. 74. 4

Answer» Correct Answer - Option 3 : 7

Ans: 3

Given:

5x2y6z is divisible by 7, 11 and 13

Concept used:

7 × 11 × 13 = 1001, when any 3 digit number multiplied by 1001 then it repeats itself and always completely divisible by 7,11 and 13.

Calculation:

Let 3 digit number be pqr.

⇒ pqr × 1001 = 5x2y6z

⇒ pqrpqr = 5x2y6z

Compare both sides

⇒ p = 5, q = 6 and r = 2

Hence, that number is 562562.

⇒ 5x2y6z = 562562

by comparing both numbers, x = 6, y = 5, and z = 2

Required value = (x – y + 3z)

⇒ 6 – 5 + 3 × 2

⇒ 7

∴ The required value is 7.

Discussion

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