Question

if x, y, z be the digit of a number beginning from the left, the number is

Answer 1

hii dear!!!
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▪in middle Z + Y = Z shows a carry so, Z + Y + 1 = 10 + Z --> Y = 9 , X + X + 1 = 9 --> X = 4 so, Z = 5 ,459 + 495 = 954
7
X = 4
Y= 9
Z = 5

______________________________________hope help u dear plz mark brainlist friend☺☺

Answer 2

The answer is given below :

Any three digit number is 123.

Now, 123

= 1×100 + 2×10 + 3×1

The value of any digit at ones place is
(1×that digit),

The value of any digit at tens place is
(10×that digit) and

The value of any digit at hundreds place is
(100×that digit).

In this problem, the digits of the number from the left are x, y and z respectively.

So, x is at hundreds place, y is at tens place and z is at ones place.

Thus, the required number be

= (100×x) + (10×y) + (1×z)

= (100x + 10y + z) [Answer]

Thank you for your question.