Find the number of integers in the range from 1 to N that ends with 2

Question

Task

As input you have:

• a positive integer N

And you should output:

• The number of integers in \$[1,N]\$ (an inclusive range) which end with the digit \$2\$ in base ten.

Test cases

1 -> 0
2 -> 1
5 -> 1
10 -> 1
12 -> 2
20 -> 2
30 -> 3
54 -> 6
97 -> 10
100 -> 10

Rules

It is a code-golf so the lowest score in bytes wins!

Answer 1

Python 2, 17 bytes

lambda n:(n+8)/10

Try it online! Uses Python 2's integer division. In Python 3 would be a byte longer with lambda n:(n+8)//10.

Answer 2

Jelly, 4 bytes

Ḋm⁵L

A monadic Link accepting a positive integer, N, which yields a non-negative integer.

Try it online!

How?

Every tenth number starting with \$2\$ ends with the digit \$2\$...

Ḋm⁵L - Link: integer, N                           e.g. 15
Ḋ    - dequeue (implicit range [1..N]) -> [2..N]       [2,3,4,5,6,7,8,9,10,11,12,13,14,15]
  ⁵  - literal ten                                     10
 m   - modulo slice                                    [2,12]
   L - length                                          2

---

Alternative 4 byter:

+8:⁵

Add eight, integer divide by ten (as first used in RGS's Python answer I believe).

Answer 3

Perl 5 -pl, 13 10 bytes

@Grimmy got it back down to 10 bytes, with correct output.

$_+=1<chop

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Answer 4

APL (Dyalog Unicode), 8 7 bytes

-1 byte thanks to Bubbler

Full program

⌊.1×8+⎕

Try it online!

Answer 5

Jelly, 7 6 5 bytes

R%⁵ċ2

Try it online! Thanks to Nick Kennedy for saving me one byte.

How it works:

R      Range from 1 to n,
 %⁵    modulo 10.
   ċ2  Then count how many of those are 2.

Answer 6

05AB1E, 4 bytes

8+T÷

Same approach as everyone else.

Try it online or verify all test cases.

Some (slightly) more interesting 5-bytes alternatives:

LT%2¢
L€θ2¢
L2Å¿O
FNθΘO

Try each online.

Explanation:

8+     # Add 8 to the (implicit) input-integer
  T÷   # Integer-divide it by 10
       # (after which the result is output implicitly)

L      # Push a list in the range [1, (implicit) input-integer]
 T%    # Take modulo-10 on each
       # or
 €θ    # Leave the last digit of each
   2¢  # Count the amount of 2s
       # (after which the result is output implicitly)

L      # Push a list in the range [1, (implicit) input-integer]
 2Å¿   # Check for each whether it ends with a 2 (I'm actually surprised it vectorizes..)
    O  # Sum to get the amount of truthy values

F      # Loop `N` in the range [0, (implicit) input-integer):
 N     #  Push `N`
  θ    #  Pop and leave only its last digit
   Θ   #  05AB1E trutify: check if it's exactly 1
    O  #  Sum all values on the stack together
       # (after the loop, the result is output implicitly)

Answer 7

Retina 0.8.2, 15 13 bytes

.+
$*
.{2,10}

Try it online! Edit: Saved 2 bytes thanks to @Grimmy. Explanation:

.+
$*

Convert to unary.

.{2,10}

Count the number of multiples of 10, each of which contains an integer that ends in 2 in base 10, plus count an additional match for a final 2-9, as that's enough for one last integer that ends in 2 in base 10.

Answer 8

Desmos

I know I'm 7 months late but this is my first code golf answer. I'm looking for some easier coding challenges to take down. I have two answers(one where I tried it without looking at any answers, then one after I looked through some answers.).

First Answer, 62 bytes:

f(N)=\sum_{n=1}^N\left\{\operatorname{mod}(n,10)=2:1,0\right\}

Try it on Desmos!

Explanation:

f(N)=                                               a function taking in an argument of N
     \sum_{n=1}^N                                   summation from 1 to N
                 \left\{                            starting piecewise
                        \operatorname{mod}(n,10)=2: if the remainder of n/10 is 2...
1                                                   sum 1
 ,                                                  otherwise...
  0                                                 sum 0
   \right\}                                         end piecewise

Not too sure why I can't take out the \left and \right for the brackets({ and }). Theoretically it should work(I've taken out the \left's and \right's of all the other "left-right pairs"), but I guess Desmos doesn't allow it.

Second Answer, 20 18 bytes:

Saved two bytes thanks to @golf69

f(N)=floor(.1N+.8)

Try it on Desmos!

Explanation:

My answer is equivalent to f(N)=floor((N+8)/10), which is explained in RGS's comment under his answer.

Answer 9

Python 3, 55 40 bytes

Any help with golfing it would be appreciated

lambda N:sum(x%10==2for x in range(N+1))

Lots of thanks to the commentors for helping me out( I have problems with memory, see names below)

Answer 10

Haskell, 14 Bytes

f x=div(x+8)10

Answer 11

W, 4 bytes

8+T/

Just another boring formula: Add eight, divide by 10. (W performs integer division if both operands are integers.)

W d, 5 bytes

[ⁿNy|

Uncompressed:

Tm2=Wk

Explanation

    W % For every number in the range [1 .. N]:
      % Keep all that satisfies:
Tm    % After modulo by 10,
  2=  % The result is equal to 2
     k% Find the length of that

Answer 12

Powershell, 50 20 bytes

1.."$args"-match"2$"

$args are arguments to pass as number.

Answer 13

Julia 1.0, 12 bytes

x->(x+8)÷10

Try it online!

Answer 14

Kotlin, 25 bytes

{(1..it).count{it%10==2}}

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Answer 15

Pyt ^[sic!], 3 bytes

Finally found the right language. I had a now-deleted answer in Vim, but it returned the empty string for an input of 1 :(

8+₀

Explanation:

8    In fact, I have no idea whether is this language stack-based, I guess it pushes 8
 +   add that 8 to the seemingly-implicit input
  ₀  divide by 10. There are also instructions to divide by numbers from 2 to 11 :)

Try it online!

Answer 16

Husk, 7 bytes

#ȯ=2→dḣ

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Answer 17

Japt -x, 6 bytes

The straightforward solution of adding 8 and floor dividing by 10 would be a byte shorter: +8 zA. But where's the fun in that?!

õ_ì̶2

Try it

õ_ì̶2     :Implicit input of integer U
õ          :Range [1,U]
 _         :Map
  ì        :  To digit array
   Ì       :  Last element
    ¶2     :  Is equal to 2?
           :Implicit output of sum of resulting array

Answer 18

Jelly, 10 7 bytes

ṾṪ=”2)S

Try it online!

But there's already 2 shorter answers I hear you say... Well, this doesn't use mathematics but rather uses strings

Answer 19

Scala, 11 bytes

n=>(n+8)/10

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Answer 20

Husk, 5 4 bytes

Edit: -1 byte thanks to Jo King

hs+8

Try it online! or check all test cases

How?

            # implicit input
    +8      # plus 8            
   s        # convert to string
  h         # remove last character
            # (so hs effectively divides by 10)

Answer 21

C (gcc), 17 bytes

f(n){n=(n+8)/10;}

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Answer 22

C (gcc), 17 bytes

f(n){n+=8;n/=10;}

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Alternative 17-byter:

C (gcc), 17 bytes

f(n){n=n/10.+.8;}

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Alternative 17-byter

C (gcc), 17 bytes

f(n){n=(n+8)/10;}

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Answer 23

Clojure, 28 bytes

(defn e[n](int(/(+ n 8)10)))

Ungolfed:

(defn ends-in-two [n]
  (int (/ (+ n 8) 10)))

Test harness:

(println (e 1))
(println (e 2))
(println (e 5))
(println (e 10))
(println (e 12))
(println (e 20))
(println (e 30))
(println (e 54))
(println (e 97))
(println (e 100))

Try it online!

Answer 24

GolfScript, 6 bytes

GolfScript has no decimal support, that's why the / works.

~8+10/

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Answer 25

Pyth, 5 bytes

/+8QT

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Explanation

/+8QT
   Q   : Variable containing evaluated input
 +8    : Add 8 to it
/   T  : Divide result of add by 10

Answer 26

Javascript (node) - 25 Bytes

f=n=>n?(n%10==2)+f(n-1):0

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Answer 27

JavaScript (V8), 15 bytes

f=n=>(n+8)/10|0

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Answer 28

PHP, 18 17 bytes

<?=$argn/10+.8|0;

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-1 bytes thanks to @oxgeba

Answer 29

Python 3, 22 18 bytes

lambda n:(n+8)//10

Thanks to @JoKing for suggesting to use integer division.

Answer 30

R, 18 15 bytes

Thanks @Giuseppe! Guess I didn't really know what the %/% operator did.

(scan()+8)%/%10

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