In a 3 digit number, the tens digit equals to the sum of the hundreds digit and the units digit. The hundreds digit of this number is 3 more than the units digit. If the digits were reserved, the new number is 11 less than half of the original number. Find the original number.

The original number is 572.

Step-by-step explanation:

Let the 3 digit number by xyz.

Where x is the hundreds digit, y is the tens digit and z is the units digit.

Hence, the digit is (100x + 10y + z).

From the question, the tens digit equals to the sum of the hundreds digit and the units digit.

That is,

y = x + z (1)

Also, from the question, the hundreds digit of this number is 3 more than the units digit.

That is,

x = z + 3 (2)

Also, from the question, If the digits were reserved, the new number is 11 less than half of the original number

If the digit were reversed, that is xyz becomes zyx. Then, the new number is (100z + 10y + x).

the new number is 11 less than half of the original number

That is,

(100z + 10y + x) + 11 = 1/2(100x + 10y + z)

This becomes

200z + 20y + 2x + 22 = 100x + 10y + z

200z - z +20y -10y +2x - 100x = -22

199z + 10y -98x = -22 (3)

Solving the equations simultaneously,

Put the value of x in equation (2) into equation (1)

x = z + 3 (2)

y = x + z (1)

That is,

y = (z+3) + z

y = 2z + 3 (4)

Put the value of y in equation (4) into equation (3)

199z + 10y -98x = -22

199z + 10(2z +3) -98x = -22

199z + 20z + 30 -98x = -22

219z -98x = -22-30

219z -98x = -52 (6)

Put the value of x in equation (2) into equation (6)

219z -98(z+3) = -52

219z -98z -294 = -52

121z = -52 + 294

121z= 242

z = 242/121

∴ z = 2

Put the value of z into equation (2)

x = z + 3

x = 2 + 3

∴ x = 5

Put the values of z and x into equation (1)

y = x + z

y = 5 + 2

∴ y = 7

Hence, x = 5, y = 7 and z = 2.

Recall that the original number is 100x + 10y + z.

100x + 10y + z = 100(5) + 10(7) +2 = 500 + 70 + 2 = 572

Hence, the original number is 572.

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)

a rotation about point l

step-by-step explanation:

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