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How Many Squares 1² to 1999² Have an Odd Tens Digit | IOQM

JEE Maths question with a full step-by-step solution.

Question

How many of the numbers

1^2, 2^2, 3^2, \ldots, 1999^2

have an odd digit in the tens place?

Solution

Answer: 400

Step 1: Write

n = 10a + b

with

a \ge 0

and

b \in \{0, 1, 2, \ldots, 9\}

. Then

n^2 = 100a^2 + 20ab + b^2.

Step 2: The last two digits of

n^2

come from

20ab + b^2

. Writing

b^2 = 10\,t_b + u_b

(tens digit

t_b

, units digit

u_b

), the tens digit of

n^2

is the units digit of

2ab + t_b

. Since

2ab

is even, the tens digit of

n^2

has the same parity as

t_b

, the tens digit of

b^2

. Step 3: Examine

b^2

for

b = 0, 1, \ldots, 9

0,\ 1,\ 4,\ 9,\ 16,\ 25,\ 36,\ 49,\ 64,\ 81,

with tens digits

0, 0, 0, 0, 1, 2, 3, 4, 6, 8

. The tens digit is odd only for

b = 4

(from

16

) and

b = 6

(from

36

). Step 4: Hence

n^2

has an odd tens digit precisely when

n \equiv 4

n \equiv 6 \pmod{10}

. In

1 \le n \le 1999

there are

200

values with

n \equiv 4

(namely

4, 14, \ldots, 1994

) and

200

values with

n \equiv 6

(namely

6, 16, \ldots, 1996

)$ Step 5: Total

= 200 + 200 = 400

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