If a, b, c and d denote the hundreds, tens, units and tenths digits re

IMO- A

z=100a+10b+c+d/10+3/100+6/1,000

z= abc.d36

Is the number that results when z is rounded to the nearest integer divisible by 3?

(1) When z/100 is rounded to the nearest hundredths, the result is 1.85

z/100 = a.bcd36
so a=1, b= 8 and
c= 4 (with d= 6,7,8,9) or 5 (with d= 0,1,2,3,4)
Least = 184.636 ~ 185 (rounding off)
Largest = 185.436 ~ 185 (rounding off)
Both divisible . Infact all numbers will round off to 185.

Sufficient.

(2) When z +2 is rounded to the nearest tens, the result is 190

Numbers between 185.d36 to 194.d36 when rounded off to nearest tens = 190
so, 185.d36 < z+2 < 194.d36
=> 183.d36 < z < 192.d36
rounding to nearest integer, Various possible values of Z= {183,184,................., 193}
So may/maynot be divisible

Not sufficient.

\(z=100a+10b+c+\frac{d}{10}+\frac{3}{100}+\frac{6}{1,000}\)

If a, b, c and d denote the hundreds, tens, units and tenths digits respectively in the decimal representation of z above, is the number that results when z is rounded to the nearest integer divisible by 3?

a + b + c = 3K, 0≤d≤4, where k is an integer ?? OR
a + b + c + 1 = 3K, 5≤d≤9 ??

(1) When \(\frac{z}{100}\) is rounded to the nearest hundredths, the result is 1.85
\(\frac{z}{100} = \frac{100}{100}a+\frac{10}{100}b+\frac{c}{100}+\frac{d}{1000}+\frac{3}{10000}+\frac{6}{100,000}\)
\(\frac{z}{100} = a+\frac{1}{10}b+\frac{c}{100}+\frac{d}{1000}+\frac{3}{10000}+\frac{6}{100,000}\)
\(\frac{z}{100}\) = a.bcd36

Since only hundredths rounding is applicable, \(\frac{z}{100}\) = a.bcd

1.845 ≤ a.bcd ≤ 1.849 OR 1.850 ≤ a.bcd ≤ 1.854
184.5 ≤ abc.d ≤ 184.9 OR 185.0 ≤ abc.d ≤ 185.4
a+b+c+1 = 3k NO OR a+b+c = 3k NO

SUFFICIENT.

(2) When z +2 is rounded to the nearest tens, the result is 190
185 ≤ z+2 ≤ 194
183 ≤ z ≤ 192

Case I: If z = abc.d36 = 183 or 192, YES
Case II: If z = abc.d36 = 184 or 185 or 188 NO

Since d=? not known

INSUFFICIENT.

Answer A.

Z=ABC.D36
Z rounded integer is div by 3 if sum digits is multi of 3

(1) sufic

abc.d36/100=a.bcd36 rounded(hth)=a.b(c+1,0)

d<5: r(hth)=a.b(c+0)=1.85,
r(integ)=abc=185,
sum abc=14 not div by 3

d>4: r(hth)=a.b(c+1)=1.85,
r(integ)=abc.d=184.5=185,
sum abc=14 not div by 3

(2) insufic

r(tens)=a(b+0,1)c

c<5: abc=a(b+0)0=190, abc=19(c<5)
c=2, d<5: sum abc=1+9+2 div by 3
c=4, d<5: sum abc=1+9+4 not div by 3

c>4: abc=a(b+1)0=190, abc=18(c>4)
c=6, d<5: sum abc=1+8+6 is div by 3
c=5, d<5: sum abc=1+8+5 not div by 3

ans (A)

IMO D.

Need to find if z is rounded to unit's place, ist divisible by 3 (sum of digits should be divisible by 3)

A. When z/100 is rounded to the nearest hundredths, the result is 1.85
= >>a.bcd36 rounded to hundredths = 1.85
It could be 1.845(/6/7/8/9) or 1.850(/1/2/3/4) so that rounded value would result in 1.85
So the value of z for all the above cases, when rounded to units digit, will be 185 Sufficient

B. When z +2 is rounded to the nearest tens, the result is 190
=> 190 = ab{c+2}.d36
z could be 188.036, 188.136, 188.236, 188.336, 188.436. OR. 187.536, 188.636, 188.736, 188.836, 188.936

In all the above cases when z+2 is rounded to nearest tens, the result is 190

So the value of z for all the above cases, when rounded to units digit, will be 188 SUFFICIENT