This is the quickest way: 1. The number must be ending with 5 as original number is ending with 25 2. by inspection it is a 5 digit number begining with 3 and ending with 5. That is 3xyz5. 3. The number barred the 25 (ie 11112222) must be a product of two consecutive numbers. 4. 11112222=1111x10002 (easy to see from pattern) = 1111x3334x3=3333x3334 5. The number = 33335 (rule of square of a number ending with 5)
Menurut saya, untuk akar kwadrat itu lebih mudah dan lebih sederhana kalau DIHITUNG LANGSUNG saja dengan menerapkan bahwa (a + b)² = a² + 2ab + b² = a² + b(2a +b). Dengan mengambil setiap tahap 2 digit dihitung dari depan kebelakang. Dengan cara itu bahkan bisa untuk menghitung angka pecahan desimal.
Actually there's a pattern with n number of 3s and 5 Like 35² is One 1 two 2 1225 335²= Since there are two threes we have two ones and three twos 112225 And here we have four ones So four threes 33335 Done! Like if useful
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This is the quickest way:
1. The number must be ending with 5 as original number is ending with 25
2. by inspection it is a 5 digit number begining with 3 and ending with 5. That is 3xyz5.
3. The number barred the 25 (ie 11112222) must be a product of two consecutive numbers.
4. 11112222=1111x10002 (easy to see from pattern) = 1111x3334x3=3333x3334
5. The number = 33335 (rule of square of a number ending with 5)
by inspection ,the square near 11 is 9 ,the last digit ends at 5 must be ,so the answer is 33335. Easy.
Did you know from the beginning that it could be applied to the formula for squaring (a+b)?
1111222225 = 1111222200 + 25
= 11110 * 100020 + 25
= 11110 * （99990 + 30） + 25
= a * (9a + 30) + 25
= 9a^2 + 30a + 25
= (3a + 5)^2
= 33335^2
Menurut saya, untuk akar kwadrat itu lebih mudah dan lebih sederhana kalau DIHITUNG LANGSUNG saja dengan menerapkan bahwa (a + b)² = a² + 2ab + b² = a² + b(2a +b). Dengan mengambil setiap tahap 2 digit dihitung dari depan kebelakang. Dengan cara itu bahkan bisa untuk menghitung angka pecahan desimal.
Actually there's a pattern with n number of 3s and 5
Like 35² is
One 1 two 2
1225
335²=
Since there are two threes we have two ones and three twos
112225
And here we have four ones
So four threes 33335
Done!
Like if useful
let m=11111, 10^5=99999+1=9m+1
1111222225=(m+1)*(9m+1)+2m+3=9m^2+12m+4=(3m+2)^2
(1111222225)^(1/2)=3m+2=33333+2=33335
I did that shit in my head, dunce.
33335
Not fit for compitative exams
my calculator can do this in 3 seconds
@@jetstream-h6p ever heard of efficency, kinda what evolution banks on
@@jetstream-h6p so how about you keep your words to yourself and grow up
This is a maths problem, not an arithmetic problem.
@@brodieanderson9598 Why don't you use chatGPT to comment on this page. It is not a sentence,but just an enume ration of words.
You should get a faster calculator.