The sum of all the numbers formed by the digits x, y and z of the number xyz is divisible by
(a) 11
(b) 33
(c) 37
(d) 74
Solution:
xyz can be written as yxz, and yzx
xyz = x × 10² + y × 10¹ + z × 10⁰
yxz = y × 10² + x × 10¹ + z × 10⁰
zyx = z × 10² + y × 10¹ + x × 10⁰
Adding all the three numbers we get:
xyz + yxz + zyx = 111x + 111y + 111z = 111(x + y + z)
Since 111 is divisible by 37 we can say that xyz + yxz + yzx is divisible by 37
The correct choice is (c )
✦ Try This: xyz - yxz - yzx is divisible by (a) 11, (b) 33, (c) 37, (d) 89
xyz = x × 10² + y × 10¹ + z × 10⁰
yxz = y × 10² + x × 10¹ + z × 10⁰
yzx = y × 10² + z × 10¹ + x × 10⁰
Therefore:
xyz - yxz - zyx = x × 10² + y × 10¹ + z × 10⁰ - (y × 10² + x × 10¹ + z × 10⁰) - (z × 10² + y × 10¹ + x × 10⁰)
xyz - yxz - zyx = 89x - 89y - 89z = 89(x - y - z)
Therefore xyz - yxz - yzx is divisible by 89.
The correct answer is the choice (d).
☛ Also Check: NCERT Solutions for Class 8 Maths Chapter 16
NCERT Exemplar Class 8 Maths Chapter 13 Problem 5
The sum of all the numbers formed by the digits x, y and z of the number xyz is divisible by (a) 11, (b) 33, (c) 37, (d) 74
Summary:
The sum of all the numbers formed by the digits x, y and z of the number xyz is divisible by 37
☛ Related Questions:
- A four-digit number aabb is divisible by 55. Then possible value(s) of b is/are (a) 0 and 2, (b) 2 a . . . .
- Let abc be a three digit number. Then abc + bca + cab is not divisible by (a) a + b + c, (b) 3, (c) . . . .
- A four-digit number 4ab5 is divisible by 55. Then the value of b - a is (a) 0, (b) 1, (c) 4, (d) 5
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