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I still remember the time when I was in high school, when I was not allowed to use calculator in tests (mathematics, physics, accountancy, etc.). A thing that was very annoying is that when I knew how to do some questions and I had done them well, but the marks were not perfect just because I made mistakes in calculating. Moreover, for the multiple choice questions, the marks will be zero if wrongly answered. Troublesome.

Aku masih ingat masa SMP-SMU, ketika tak diperbolehkan pakai kalkulator saat ulangan (matematika, fisika, akuntansi, dsb). Hal yang sangat mengesalkan yaitu waktu soal-soal tau cara mengerjakannya dan sukses dikerjakan, tapi nilainya tak sempurna hanya karena salah hitung. Malah untuk soal pilihan ganda, nilai akan nol kalau salah jawab. Repot.

Fortunately when I was in primary school I read a book whose title I don't remember, about how to check calculation correctness. It is applicable for addition, subtraction, multiplication and division without remainder. The method is also very simple, and the time needed to check is less than 1/4 of time to calculate.

Untungnya waktu SD aku baca suatu buku yang entah apa judulnya lupa, mengenai cara mengecek kebenaran hitungan. Berlaku untuk tambah, kurang, kali, dan bagi tanpa sisa. Caranya pun mudah, dan waktu yang diperlukan untuk mengecek kurang dari 1/4 waktu menghitung.

Checksum

First we need to learn checksum (a sum for checking) of a number. How to do it: take the sum of every digit of the number, and if the result is more than one digit, take the sum of each digit again, and so on until you get a single digit. For example:

Checksum

Pertama kita perlu belajar checksum (jumlah untuk mengecek) suatu bilangan. Caranya, jumlahkan setiap angka dari bilangan itu, dan jika hasilnya lebih dari satu angka, jumlahkan lagi tiap angkanya, dan seterusnya sampai didapat satu angka. Contoh:

  • 7 → 7
  • 53 → 5+3 = 8
  • 456 → 4+5+6 = 15 → 1+5 = 6
  • 6788 → 6+7+8+8 = 29 → 2+9 = 11 → 1+1 = 2

For exercise (just to make sure that you understand before we continue), what is the checksum of the following numbers? Click Check to verify your answers.

  • 278 → checksum =
  • 9999 → checksum =

Untuk latihan (untuk memastikan Anda mengerti sebelum kita lanjutkan), berapa checksum bilangan berikut? Klik Cek untuk periksa jawaban.

  • 368 → checksum =
  • 99198 → checksum =

How to Check

As an example we start from addition. This is the rule. If the checksum of sum of the checksum of the first number and the checksum of the second number is different from the checksum of the sum of the first number and the second number, the calculation is surely wrong. If it's the same, it's not surely correct but most probably correct.

Are you confused? If yes, then you are normal. If no, then you are also normal.

Do you need shorter notation? Checksum of a number x is called c(x), so: if c(c(a) + c(b)) ≠ c(a+b), the calculation is wrong, otherwise, the calculation is likely to be correct.

Cara Periksa

Sebagai contoh kita mulai dari penambahan. Begini peraturannya. Jika checksum jumlah checksum bilangan pertama dan checksum bilangan kedua beda dengan checksum jumlah bilangan pertama dan bilangan kedua, pasti hitungannya salah. Jika sama, belum tentu betul tapi kemungkinan besar betul.

Bingungkah? Jika iya, Anda normal. Jika tidak, Anda juga normal.

Perlu notasi lebih singkat? Checksum suatu bilangan x kita sebut c(x), maka: jika c(c(a) + c(b)) ≠ c(a+b), hitungan salah, jika sebaliknya, hitungan hampir pasti betul.

Real Examples

      Checksum After we calculate c(75) = 3 and c(34) = 7, we calculate 3+7 = 10, then we take the checksum of it again, c(10) = 1. Then we compare it with c(119), which is 1. It's the same! So, it's likely that the calculation is correct (and it's really correct)
  7 5 3
  3 4 7
+
 
1 0 9 1

          Checksum c(7643) = 2, c(3901) = 4, so c(7643+3901) should be 2+4 = 6, right? But when we calculate c(10544) the result is not 6, but 5! So the calculation is wrong! (should be 11544)
  7 6 4 3 2
  3 9 0 1 4
+
 
1 0 5 4 4 5

Contoh Nyata

      Checksum Setelah kita hitung c(75) = 3 dan c(34) = 7, kita hitung 3+7 = 10, lalu ambil checksumnya lagi, c(10) = 1. Bandingkan dengan c(119), yaitu 1. Wah sama! Maka, kemungkinan besar hitungannya betul (dan memang betul)
  7 5 3
  3 4 7

+
1 0 9 1

          Checksum c(7643) = 2, c(3901) = 4, jadi c(7643+3901) harusnya 2+4 = 6, kan? Tapi waktu kita hitung c(10544) hasilnya bukan 6, tapi 5! Jadi hitungannya salah! (harusnya 11544)
  7 6 4 3 2
  3 9 0 1 4

+
1 0 5 4 4 5

Other Operations

How about the other operations? For subtraction, just reverse the addition above. So for a - b = h, verify that c(c(h) + c(b)) = c(a). For multiplication, the method is the same as the addition, that is for a × b = h, verify that c(c(a) × c(b)) = c(h). And for division? Reverse multiplication.

Operasi Lain

Gimana dengan operasi lain? Untuk pengurangan, balik saja penambahan di atas. Jadi untuk a - b = h, pastikan bahwa c(c(h) + c(b)) = c(a). Untuk perkalian, sama dengan penambahan caranya, yaitu untuk a × b = h, pastikan bahwa c(c(a) × c(b)) = c(h). Dan untuk pembagian? Balik perkalian.

Calculating the Checksum Itself Takes Too Much Time!

Yes, it is. But of course there are methods to make it faster. That is, discard two or more digits that sum to 9, after that calculate the checksum. Example:

  • What is the checksum of 8517 ?
  • You can see that 8 and 1 have sum of 9 (the more you do this the more you become expert on this), so we discard 8 and 1. Now what remains is 57.
  • Calculate the checksum of 57, which is 3.

More than two digits also can be discarded, as long as the sum is 9 or multiples of 9 (18, 27, etc). Furthermore, 0 and 9 itself can be discarded. Example:

  • What is the checksum of 414020193 ?
  • Discard every instance of 0 and 9, so it becomes 414213
  • 4, 1 and 4 have sum of 9, discard them too, so now it is 213
  • Calculate the checksum of 213, which is 6.

Hitung Checksumnya Aja Buang Banyak Waktu!

Memang. Tapi tentu ada cara mempercepatnya. Yaitu, buanglah dua atau lebih angka yang jumlahnya 9, baru hitung checksumnya. Contoh:

  • Berapa checksum 8517 ?
  • Dapat dilihat 8 dan 1 jumlahnya 9 (semakin terbiasa semakin ahli), jadi kita buang 8 dan 1 dari 8517, tinggal 57.
  • Hitung checksum 57, yaitu 3.

Lebih dari 2 angka pun bisa, asal jumlahnya 9, atau kelipatan 9 (18, 27, dsb). Sedangkan 0 dan 9 sendiri tentu saja dibuang. Contoh:

  • Berapa checksum 414020193 ?
  • Buang setiap 0 dan 9, jadi tinggal 414213
  • 4, 1, dan 4 dijumlah hasilnya 9, buang juga, jadi tinggal 213
  • Hitung checksum 213, yaitu 6.

Interactive Test

Use the form below to try the method mentioned above. Fill in the numbers that you want to calculate in the 3 pink boxes below, and you will be told whether your answer is correct or not depending on the checksum.

Tes Interaktif

Pakailah isian di bawah ini untuk coba mempraktekkan cara di atas. Isi angka yang ingin Anda hitung di 3 kotak merah muda, dan Anda akan diberitahu apakah jawaban Anda benar atau salah berdasarkan checksum.

 

Written by: yuku

wah.....agak aneh juga ya, tapi terobosannya boleh juga! Penyajiannya keren juga bisa dicobain :P

Derianto Kusuma [id], 4 Dec 2005, 14:22 reply
Huahuahua...
458 8 458 + 595 = 9, is probably correct!
595 1 parah juga.... =P
---- +
9 9


Wahyu PY [sg], 5 Dec 2005, 9:09 reply
Cool. This is the first time I've seen this kind of technique. You're surely lucky to have access to such gifted mathematics book since primary school. I wish I had, so I didn't have to recalculate 716 * 1407 several times to make sure it is correct.
Anyway, thanks for sharing. And those JavaScript extras are really nice and helpful.


budi [id], 5 Dec 2005, 11:03 reply
Wah, bisa gitu yah...
Hebaaat... ga kepikir sama sekali...


Lesmanahadi [id], 8 Dec 2005, 14:43 reply
awesome.... i never realize it at all.. thx

pank [id], 17 Mar 2006, 15:20 reply
guys, you should check out "vedict math" in the web. it gives a lot short way in math calculation. try...

win [id], 3 Sep 2006, 15:57 reply
sorry for the mistypo, it should have been "vedic math"

win [id], 3 Sep 2006, 16:01 reply
wah..bagus.. :))

den [id], 17 Sep 2006, 8:46 reply
hhe. krenn,,, bwat nambah pengetahuan
makachii ya.. :)


chii~ [id], 8 Sep 2007, 16:00 reply
amazing,now i can math with short time,thank you for your tips.

Riski [id], 15 Sep 2008, 2:22 reply
interest, semoga bisa diterapin oleh semua anak indonesia

cagur.com [id], 15 Sep 2008, 2:29 reply
bagaimana cara menghitung checksum dengan menggunakan 2's dari KTP ???

rita [--], 9 Oct 2013, 20:20 reply